The proof is analogous to the proof of Proposition <ref>. If we replace $\DNC$ everywhere with $\DNC_\pi$ (which is the statement we will need), an alternative proof can be given by showing that the embedding

\[\DNC_\pi(\algebr,\subalgebr) \ra \blup_{\pi \times \id_{\mathbb{R}}}(\algebr \times \mathbb{R}, \subalgebr \times \{0\}),\]

from Proposition <ref>, is a Lie subalgebroid. The result then follows from Proposition <ref>.

For the rest of the statements in this section, the key is to use the following lemma.

Let $\algebrtwoid$ be a Lie algebroid. The canonical diffeomorphism

\[\blup(\algebr \times \algebrtwo,\subalgebr \times \algebrtwo) \xra{\sim} \blup(\algebr,\subalgebr) \times \algebrtwo\]

(from Proposition <ref>)
restricts to an isomorphism of Lie algebroids

\[\varphi: \blup_{\pi_\algebr \times \pi_\algebrtwo}(\algebr \times \algebrtwo,\subalgebr \times \algebrtwo) \xra{\sim} \blup_{\pi_{\algebr}}(\algebr,\subalgebr) \times \algebrtwo.\]

That the restriction to $\blup_{\pi_{\algebr}}(\algebr,\subalgebr) \times \algebrtwo$ maps bijectively onto $\blup_{\pi_\algebr \times \pi_\algebrtwo}(\algebr \times \algebrtwo,\subalgebr \times \algebrtwo)$ follows from the fact that the isomorphism from Proposition <ref> is induced by the map

\[g: \DNC(\algebr \times \algebrtwo, \subalgebr \times \algebrtwo) \xra{\sim} \DNC(\algebr, \subalgebr) \times \algebrtwo \textnormal{ given by } z \mapsto
\begin{cases}
(b,\xi,c) & \text{if}\ z=(b,c,\xi,0) \\
(a,1,c) & \text{if}\ z=(a,c,t),
\end{cases}\]

and one can readily verify that $\DNC(\pi_\algebr) \circ \pr_1(g(z)) \in \subbase \times \mathbb{R}$ if and only if $\DNC(\pi_\algebr \times \pi_\algebrtwo)(z) \in \subbase \times \basetwo \times \mathbb{R}$
. Observe that, under the diffeomorphism $\varphi$,

\[(\algebr \times \algebrtwo)|_{(\base \times \basetwo) \setminus (\subbase \times \basetwo)} = (\algebr \times \algebrtwo)|_{(\base \setminus \subbase) \times \basetwo}\]

maps bijectively onto $\algebr|_{\base \setminus \subbase} \times \algebrtwo$. Now the result follows: since $\blup_{\pi_\algebr \times \pi_\algebrtwo}(\algebr \times \algebrtwo, \subalgebr \times
\algebrtwo)$ has a unique Lie algebroid structure for which $(\algebr \times \algebrtwo)|_{(\base \setminus \subbase) \times \basetwo}$ is a Lie subalgebroid, we see that the Lie algebroid structures of $\blup_{\pi_\algebr \times \pi_\algebrtwo}(\algebr \times \algebrtwo, \subalgebr \times
\algebrtwo)$ and $\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \algebrtwo$ have to agree (under the map $\varphi$). This proves the statement.

Notice that a similar argument shows that we can replace “$\blup$” everywhere with “$\DNC$”, and we even have an isomorphism of Lie algebroids $\DNC(\algebr \times \algebrtwo, \subalgebr \times \algebrtwo) \xra{\sim} \DNC(\algebr,\subalgebr) \times \algebrtwo$.

We start with showing that the blow-down map for blow-ups of Lie algebroids becomes a morphism of Lie algebroids.

The blow-down map $\bldown: \blup_\pi(\algebr,\subalgebr) \ra \algebr$ is a morphism of Lie algebroids.

We denote the blow-down map on the base manifold by $\bldown_\base$.

By Corollary <ref>, we only have to check that
$\gr \bldown \subset \blup_\pi(\algebr,\subalgebr) \times \algebr$ is a Lie subalgebroid. By Lemma <ref>, there is a canonical isomorphism of Lie algebroids

\begin{equation}\label{eq: iso blups in blow-down map LA}
\blup_\pi(\algebr,\subalgebr) \times \algebr \xra{\sim} \blup_{\pi \times \pi}(\algebr \times \algebr, \subalgebr \times \algebr).
\end{equation}

Notice that it restricts to an isomorphism of Lie algebroids

\[\mathbb{P}_\pi(\algebr,\subalgebr) \times \algebr \xra{\sim} \mathbb{P}_{\pi \times \pi}(\algebr \times \algebr, \subalgebr \times \algebr),\]

so, by explicitly writing down what elements of $\gr \bldown$ are mapped to under (<ref>), we see that, under (<ref>), $\gr \bldown \subset \blup_\pi(\algebr,\subalgebr) \times \algebr$ maps bijectively onto $\blup_{(\pi \times \pi)|_{\Delta_\algebr}}(\Delta_\algebr,\Delta_\subalgebr) \subset \blup_{\pi \times \pi}(\algebr \times \algebr,\subalgebr \times \algebr)$:

\begin{align*}
(z,p(z)) =
\begin{cases}
([b,\xi],b) \\
\end{cases} \mapsto
\begin{cases}
[b,b,\xi,0] & \text{if}\ z=[b,\xi] \\
[a,a,1] & \text{if}\ z=[a,1].
\end{cases}
\end{align*}

We can now equivalently prove that

\[\blup_{(\pi \times \pi)|_{\Delta_\algebr}}(\Delta_\algebr,\Delta_\subalgebr) \subset \blup_{\pi \times \pi}(\algebr \times \algebr, \subalgebr \times \algebr)\]

is a Lie subalgebroid. Since $\Delta_\algebr \cap (\subalgebr \times \algebr) = \Delta_\subalgebr$, the result follows from Proposition <ref>.

Alternatively, the above statement is a consequence of the fact that $\bldown: \blup_{\pi}(\algebr,\subalgebr) \ra \algebr$ is an isomorphism on a dense open subset of $\blup_{\pi}(\algebr,\subalgebr)$.

We now prove the following analogue of Proposition <ref>.

Let $\varphi: (\algebr,\subalgebr) \ra (\algebrtwo,\subalgebrtwo)$ be a morphism of Lie algebroids between pairs of Lie algebroids $(\algebroid,\subalgebroid)$ and $(\algebrtwoid,\subalgebrtwoid)$. If the base map $f: (\base,\subbase) \ra (\basetwo,\subbasetwo)$ of $\varphi$ satisfies $f^{-1}(\subbasetwo)=\subbase$ and $d_\normal f$ is fiberwise injective, then $\blup(\varphi)$ restricts to a morphism of Lie algebroids $\blup_{\pi_\algebr}(\algebr,\subalgebr) \ra \blup_{\pi_{\algebrtwo}}(\algebrtwo,\subalgebrtwo)$.

By Proposition <ref>, we only have to check that $\gr \blup(\varphi)$ is a Lie subalgebroid of $\blup_{\pi_\algebr}(\algebr, \subalgebr) \times \blup_{\pi_\algebrtwo}(\algebrtwo,\subalgebrtwo)$. First of all, by Lemma <ref>, there is an isomorphism of Lie algebroids

\begin{equation}\label{eq: iso blups in LA morphism to blup}
\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \blup_{\pi_{\algebrtwo}}(\algebrtwo,\subalgebrtwo) \xra{\sim} \blup_{\blup(\pi_\algebr) \times \pi_\algebrtwo}(\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \algebrtwo, \blup_{\pi_\algebr}(\algebr,\subalgebr) \times \subalgebrtwo).
\end{equation}

Notice that it restricts to an isomorphism of Lie algebroids

\[\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \mathbb{P}_{\pi_\algebrtwo}(\algebrtwo,\subalgebrtwo) \xra{\sim} \mathbb{P}_{\blup(\pi_\algebr) \times \pi_\algebrtwo}(\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \algebrtwo,\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \subalgebrtwo).\]

Now, under (<ref>), $\gr \varphi$ maps bijectively onto $\blup_{(\blup(\pi_\algebr) \times \pi_\algebrtwo)|_{\gr \varphi \circ \bldown}}(\gr \varphi \circ \bldown, \gr \varphi \circ \bldown|_{\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)})$:

\[(z,\blup(\varphi)(z)) =
\begin{cases}
([b,\xi],[\varphi(b),d_\normal\varphi(b)\xi]) \\
\end{cases} \mapsto
\begin{cases}
[[b,\xi],\varphi(b),d_\normal\varphi(b)\xi] & \text{if}\ z=[b,\xi] \\
[[a,1],\varphi(a),1] & \text{if}\ z=[a,1],
\end{cases}\]

where we used that there is a canonical diffeomorphism

\[\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr) \cong \mathbb{P}_{(\blup(\pi_\algebr) \times \pi_\algebrtwo)|_{\gr \varphi \circ \bldown}}(\gr \varphi \circ \bldown, \gr \varphi \circ \bldown|_{\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)})\]

by the sequence of canonical diffeomorphisms (note: the result is not a morphism of vector bundles)

\begin{align*}
\normal_{\pi_\algebr}(\algebr,\subalgebr) &\cong (\normal_{\pi_\algebr}(\algebr,\subalgebr) \times \mathbb{R})/\mathbb{R}^\times \\
&\cong \normal_{\DNC(\pi_\algebr)}(\DNC_{\pi_\algebr}(\algebr,\subalgebr),\normal_{\pi_\algebr}(\algebr,\subalgebr))/\mathbb{R}^\times \\
&\cong \normal_{\blup(\pi_\algebr)}(\blup_{\pi_\algebr}(\algebr,\subalgebr),\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)) \cong \normal_{(\blup(\pi_\algebr) \times \pi_\algebrtwo)|_{\gr \varphi \circ \bldown}}(\gr \varphi \circ \bldown, \gr \varphi \circ \bldown|_{\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)}).
\end{align*}

where we used the canonical diffeomorphism $\normal(\DNC_\pi(\algebr,\subalgebr),\normal_\pi(\algebr,\subalgebr)) \cong \normal_\pi(\algebr,\subalgebr) \times \mathbb{R}$ (see Remark <ref>) in the second diffeomorphism, and we used Proposition <ref> in the third diffeomorphism.

We can now equivalently show that

\[\blup_{\blup(\pi_\algebr) \times \pi_\algebrtwo}(\gr \varphi \circ \bldown, \gr \varphi \circ \bldown|_{\mathbb{P}(\algebr,\subalgebr)}) \subset \blup_{\blup(\pi_\algebr) \times \pi_\algebrtwo}(\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \algebrtwo, \blup_{\pi_\algebr}(\algebr,\subalgebr) \times \subalgebrtwo)\]

is a Lie subalgebroid. And indeed, since $\bldown: \blup_{\pi_\algebr}(\algebr,\subalgebr) \ra \algebr$ is a morphism of Lie algebroids (see Corollary <ref>), so are $\varphi \circ \bldown$ and $\varphi \circ \bldown|_{\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)}$. This shows that $\gr \varphi \circ \bldown \subset \blup_{\pi_\algebr}(\algebr,\subalgebr) \times \algebrtwo$ and $\gr \varphi \circ \bldown|_{\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)} \subset \mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr) \times \subalgebrtwo$ are closed Lie subalgebroids with

\[\gr \varphi \circ \bldown \cap (\blup_{\pi_\algebr}(\algebr,\subalgebr) \times \subalgebrtwo) = \gr \varphi \circ \bldown|_{\mathbb{P}_{\pi_\algebr}(\algebr,\subalgebr)}.\]

By Proposition <ref>, the result follows.

Again, following a similar argument shows that we can replace $\blup_{\pi_\algebr}$ and $\blup_{\pi_\algebrtwo}$ everywhere with $\DNC$ (or $\DNC_{\pi_\algebr}$ and $\DNC_{\pi_\subalgebr}$, respectively):

Let $\varphi: (\algebr,\subalgebr) \ra (\algebrtwo,\subalgebrtwo)$ be a morphism of Lie algebroids between pairs of Lie algebroids $(\algebroid,\subalgebroid)$ and $(\algebrtwoid,\subalgebrtwoid)$. Then $\DNC(\varphi): \DNC(\algebr,\subalgebr) \ra \DNC(\algebrtwo,\subalgebrtwo)$ is a morphism of Lie algebroids.

We can relate the Lie algebroids $\DNC_{\pi}(\algebr,\subalgebr)$ and $\blup_{\pi}(\algebr,\subalgebr)$ as follows:

The quotient map $\DNC(\algebr,\subalgebr) \setminus \subalgebr \times \mathbb{R} \rightarrowdbl \blup(\algebr,\subalgebr)$ restricts to a morphism of Lie algebroids

\[\DNC_\pi(\algebr,\subalgebr) \rightarrowdbl \blup_\pi(\algebr,\subalgebr)\]

over the quotient map  $\DNC(\base,\subbase) \setminus \subbase \times \mathbb{R} \rightarrowdbl \blup(\base,\subbase)$.

Denote the quotient map $\DNC_\pi(\algebr,\subalgebr) \rightarrowdbl \blup_\pi(\algebr,\subalgebr)$ by $q$. We have to show that

\[\gr q \subset \DNC_\pi(\algebr,\subalgebr) \times \blup_\pi(\algebr,\subalgebr)\]

is a Lie subalgebroid. Similar to the proof of Corollary <ref>, recall that if we have a pair of manifolds $(M,N)$, and $Z$ is a manifold, then we have a canonical diffeomorphism

\[\DNC(M,N) \times Z \xra{\sim} \DNC(M \times Z, N \times Z).\]

Again, it is readily verified that, in case $M$, $N$, and $Z$ are Lie algebroids, then it is even an isomorphism of Lie algebroids, and the restriction to $\DNC_{\pi_M}(M,N) \times Z$ is an isomorphism of Lie algebroids onto $\DNC_{\pi_M \times \pi_Z}(M \times Z, N \times Z)$.
So, we obtain an isomorphism of Lie algebroids

\[\DNC_\pi(\algebr,\subalgebr) \times \blup_\pi(\algebr,\subalgebr) \xra{\sim} \DNC_{\blup(\pi) \times \pi}(\blup_\pi(\algebr,\subalgebr) \times \algebr, \blup_\pi(\algebr,\subalgebr) \times \subalgebr).\]

Observe now that $\gr q$ is mapped bijectively onto $\DNC_{(\blup(\pi) \times \pi)|_{\gr \bldown}}(\gr \bldown, \gr \bldown|_{\mathbb{P}_\pi(\algebr,\subalgebr)})$:

\begin{align*}
(z,q(z)) =
\begin{cases}
(b,\xi,[b,\xi]) \\
\end{cases} &\mapsto
\begin{cases}
([b,\xi],b,\xi) & \text{if}\ z=(b,\xi) \\
([a,1],a,t) & \text{if}\ z = (a,t),
\end{cases}
\end{align*}

where we used that there is a canonical diffeomorphism

\[\normal_\pi(\algebr,\subalgebr) \cong \normal_{\blup(\pi)}(\blup_\pi(\algebr,\subalgebr),\mathbb{P}_\pi(\algebr,\subalgebr)) \cong \normal_{(\blup(\pi) \times \pi)|_{\gr \bldown}}(\gr \bldown, \gr \bldown|_{\mathbb{P}_\pi(\algebr,\subalgebr)}) %\textnormal{ given by } (b,\xi) \mapsto ([b,\xi],d_\normal q(b,\xi)1)
\]

(see the proof of Corollary <ref>). This shows that we can equivalently prove that

\[\DNC_{(\blup(\pi) \times \pi)|_{\gr \bldown}}(\gr \bldown, \gr \bldown|_{\mathbb{P}_\pi(\algebr,\subalgebr)}) \subset \DNC_{\blup(\pi) \times \pi}(\blup_\pi(\algebr,\subalgebr) \times \algebr,\blup_\pi(\algebr,\subalgebr) \times \subalgebr)\]

is a Lie subalgebroid. Since $\gr \bldown \cap (\blup_\pi(\algebr,\subalgebr) \times \subalgebr) = \gr \bldown|_{\mathbb{P}_\pi(\algebr,\subalgebr)}$, the result follows from Corollary <ref> and Proposition <ref>.

Lastly, we present a simple example of a blow-up of Lie algebroids. It is the analogue of Example <ref> in the Lie algebroid setting.

Recall from Proposition <ref>  (or Example <ref>) that the Lie algebroid of $\blup_{\source,\target}(\base \times \base, \subbase \times \subbase) \rra \blup(\base,\subbase)$ is

\[\blup_{\pi_{T\base}}(T\base,T\subbase) \ra \blup(\base,\subbase).\]

Also, recall from Example <ref> that, as Lie groupoids, $\blup_{\source,\target}(\base \times \base, \subbase \times \subbase)$ is isomorphic to $(P \times_{\mathbb{R}} P)/\mathbb{R}^\times \rra P/\mathbb{R}^\times = \blup(\base,\subbase)$, where $P \coloneqq \DNC(\base,\subbase) \setminus \subbase \times \mathbb{R}$, which has Lie algebroid

\[(\ker d\hat{t}|_P)/\mathbb{R}^\times \ra \blup(\base,\subbase)\]

(see Example <ref>), so we can identify $\blup_{\pi_{T\base}}(T\base,T\subbase)$ with $(\ker d\hat{t}|_P)/\mathbb{R}^\times \subset TP/\mathbb{R}^\times$.
Observe that our findings in Example <ref> (the blow-up of pair groupoids) agree with our findings for $\algebr \coloneqq \blup_{\pi_{T\base}}(T\base,T\subbase)$. Indeed,

\[\anchor: \blup_{\pi_{T\base}}(T\base,T\subbase) \ra T\blup(\base,\subbase)\]

is an isomorphism when restricted to $T\base|_{\base \setminus \subbase} \subset \blup_{\pi_{T\base}}(T\base,T\subbase)$ (onto $T\base|_{\base \setminus \subbase} \subset T\blup(\base,\subbase)$), and, for all $[y,\xi] \in \mathbb{P}(\base,\subbase)$, $\anchor([y,\xi])$ is surjective with a $1$-dimensional kernel. Therefore,

\[\mathfrak{g}_z = 0 \textnormal{ if } z \in \base\setminus\subbase; \quad \mathfrak{g}_z \cong \mathbb{R} \textnormal{ if } z \in \mathbb{P}(\base,\subbase)\]

Similarly, by Proposition <ref>, we see that the orbits are given by the connected components of $\base \setminus \subbase$ and the connected components with $\mathbb{P}(\base,\subbase)$; this is also in line with Example <ref>.

§.§ Blow-ups: different approaches

We will now describe different approaches of constructing a blow-up of manifolds, and later in this section also of constructing a blow-up of Lie groupoids and Lie algebroids. The difference lies in the fact that we do not “factor” through the deformation to the normal cone, but rather describe the blow-up directly. This direct approach to blow-ups has its advantages. For example, in the next approach to blow-up of manifolds, the local description is very easy to understand. On the other hand, describing the blow-up globally (e.g. the smooth structure), is a bit trickier.

§.§.§ Blow-ups of manifolds: a different approach

This construction is based on <cit.>.
Let $(\base,\subbase)$ be a pair of manifolds. We first restrict our attention to the case where

\[\subbase \coloneqq \mathbb{R}^p \subset \mathbb{R}^{p+q} = \mathbb{R}^n \eqqcolon \base.\]

Denote by $S^{q-1}$ the unit sphere in $\mathbb{R}^q$. Observe that on

\[\mathbb{R}^{p} \times S^{q-1} \times \mathbb{R}\]

(it is useful to think of this space as describing polar coordinates on $\mathbb{R}^n$) we have a canonical $\mathbb{Z}_2$-action given by

\[\tau \cdot (x,\theta,t) = (x,-\theta,-t).\]

The action is obviously free (and proper, because $\mathbb{Z}_2$ is finite), so we obtain a quotient manifold

\[\blupp(\mathbb{R}^n,\mathbb{R}^p) \coloneqq (\mathbb{R}^{p} \times S^{q-1} \times \mathbb{R})/\mathbb{Z}_2\]

(the subscript “pc" refers to polar coordinates). To obtain a blow-down map, observe that the map

\[\Psi_{\textnormal{pc}}: \mathbb{R}^{p} \times S^{q-1} \times \mathbb{R} \ra \mathbb{R}^{p+q} \textnormal{ given by } (x,\theta,t) \mapsto (x,t \cdot \theta)\]

is smooth and intertwines the $\mathbb{Z}_2$-actions (with trivial action on $\mathbb{R}^n)$. It therefore descends to a smooth map, which we will call the blow-down map:

\[\bldownn: \blupp(\mathbb{R}^n,\mathbb{R}^p) \ra \mathbb{R}^n.\]

Observe that if we restrict $\Psi_{\textnormal{pc}}$ to $\mathbb{R}^p \times S^{q-1} \times \mathbb{R}_{>0}$, then this map $\Psi_{\textnormal{pc}}^{> 0}$ is a diffeomorphism with inverse

\[(\Psi_{\textnormal{pc}}^{> 0})^{-1}: \mathbb{R}^{p+q} \setminus \mathbb{R}^p \ra \mathbb{R}^p \times S^{q-1} \times \mathbb{R}_{>0} \textnormal{ given by } (x,y) \mapsto (x, \frac{y}{\Vert y \Vert}, \Vert y \Vert).\]

This shows that the blow-down map $\bldownn$ restricts to a diffeomorphism

\[\bldownn^{-1}(\mathbb{R}^n \setminus \mathbb{R}^p) \ra \mathbb{R}^n \setminus \mathbb{R}^p.\]

Moreover, if we restrict $\Psi_{\textnormal{pc}}$ to $\mathbb{R}^p \times S^{q-1} \times \{0\}$, then this map maps onto $\mathbb{R}^p$, and the induced map

\[\mathbb{R}^p \times \mathbb{RP}^{q-1} \xra{\pr_1} \mathbb{R}^p\]

can be seen as the restriction of $\bldownn$ to $\bldownn^{-1}(\mathbb{R}^p)$.

Using the lemma below (compare it with Lemma <ref>), we will be able to lift transition maps of a pair of smooth manifolds $(\base,\subbase)$ to diffeomorphisms between the blow-ups of their respective domains. Then we will assemble the blow-ups of the (ranges of the) charts as a disjoint union. We will then define a relation on this space which identifies points that are related by the induced diffeomorphisms (that the transition maps define), and the resulting quotient space will be the blow-up $\blupp(\base,\subbase)$.
Let $h=(h_1,h_2): \mathbb{R}^{p+q} \ra \mathbb{R}^{k+\ell}$ be a smooth map such that $h^{-1}(\mathbb{R}^k) = \mathbb{R}^p$. Moreover, assume that $dh_2(x,0)$ is injective for all $x \in \mathbb{R}^p$ (i.e. $d_\normal h$ is fiber-wise injective). Then the map

\begin{align*}
\widetilde{h}: \mathbb{R}^p \times S^{q-1} \times \mathbb{R} &\ra \mathbb{R}^k \times S^{\ell-1} \times \mathbb{R} \textnormal{ given by } \\
(x,\theta,t) &\mapsto
\begin{cases}
(h_1(x,0),\frac{\tfrac{\partial h_2}{\partial \theta}(x,0)\theta}{\Vert \tfrac{\partial h_2}{\partial \theta}(x,0)\theta \Vert},0) & \text{if}\ t = 0 \\
(h_1(x,t\theta), \tfrac{t}{\vert t \vert}\tfrac{h_2(x,t\theta)}{\Vert h_2(x,t\theta) \Vert}, \tfrac{t}{\vert t \vert}\Vert h_2(x,t\theta) \Vert) & \text{if}\ t \neq 0
\end{cases}
\end{align*}

is smooth.

If we write $\widetilde{h} = (\widetilde{h}_1,\widetilde{h}_2,\widetilde{h}_3)$, it is obvious that $\widetilde{h}_1$ is smooth, so we only have to verify that $\widetilde{h}_2$ and $\widetilde{h}_3$ are smooth. Using that $h_2(x,0) = 0$, we have that

\[h_2(x,\theta) = \int_0^1 \frac{d}{dt}h_2(x,t\theta) dt = \int_0^1 \frac{\partial h_2}{\partial \theta}(x,t\theta)\theta dt = \frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,\theta)\theta,\]

where $r(x,\theta) \coloneqq -\textstyle\frac{\partial h_2}{\partial \theta}(x,0) + \textstyle\int_0^1\frac{\partial h_2}{\partial \theta}(x,t\theta)dt$, is a smooth map with the property that $r(x,0)=0$.

\[t^{-1}h_2(x,t\theta) = \frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta\]

(note: $r(x,0)=0$).
Observe that this shows that $\widetilde{h}_2$ is the map

\[(x,\theta,t) \mapsto \frac{t^2 \cdot (\frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta)}{|t^2| \cdot \Vert \frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta \Vert} = \frac{\frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta}{|\Vert \frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta \Vert}\]

and $\widetilde{h}_3$ is the map

\[(x,\theta,t) \mapsto t \cdot \Vert \frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta \Vert.\]

Since the norm $\Vert \cdot \Vert$, seen as a map $\mathbb{R}^d \ra \mathbb{R}$ (where $d\ge0$), is a smooth function away from $0 \in \mathbb{R}^d$, we see that it suffices to prove that the smooth map

\[\mathbb{R}^p \times S^{q-1} \times \mathbb{R} \ra \mathbb{R}^\ell \textnormal{ given by } (x,\theta,t) \mapsto \frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta\]

is nowhere vanishing. Let $(x,\theta,t) \in \mathbb{R}^p \times S^{q-1} \times \mathbb{R}$. If $t \neq 0$, then

\[\frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta = t^{-1}h_2(x,t\theta) \neq 0,\]

because $h_2^{-1}(\mathbb{R}^k) = \mathbb{R}^p$. If $t=0$, then, since $r(x,0)=0$, wee see that

\[\frac{\partial h_2}{\partial \theta}(x,0)\theta + r(x,t\theta)\theta = \frac{\partial h_2}{\partial \theta}(x,0)\theta \neq 0,\]

because $\frac{\partial h_2}{\partial \theta}(x,0)\theta$ is injective by assumption. This proves that both $\widetilde{h}_2$ and $\widetilde{h}_3$ are smooth maps, so this proves the statement.

Immediately we obtain the following two corollaries:
Let $h$ be as in Lemma <ref>. Then $h$ induces a smooth map

\[\blupp(h): \blupp(\mathbb{R}^n,\mathbb{R}^p) \ra \blupp(\mathbb{R}^{k + \ell}, \mathbb{R}^k \times \{0\}).\]

This follows from the fact that the map $\widetilde{h}$ from Lemma <ref> intertwines the $\mathbb{Z}_2$-actions.

Let $h: \mathbb{R}^{p+q} \ra \mathbb{R}^{k + \ell}$ and $h': \mathbb{R}^{k + \ell} \ra \mathbb{R}^{m+n}$ satisfy the hypothesis of Lemma <ref>. Then $\blupp(h' \circ h) = \blupp(h') \circ \blupp(h)$.

This follows from the definitions of $\widetilde{h}$ and $\widetilde{h}'$.

Before we move to the general case of blow-ups of pairs of smooth manifolds, observe the following: if $U,U' \subset \mathbb{R}^{p+q}$ are open subsets, then we can view $\blupp(U,U \cap \mathbb{R}^p) \subset \blupp(\mathbb{R}^n, \mathbb{R}^p)$ and $\blupp(U',U' \cap \mathbb{R}^p) \subset \blupp(\mathbb{R}^n, \mathbb{R}^p)$ as open subsets, and

\[\blupp(U,U \cap \mathbb{R}^p) \cap \blupp(U',U' \cap \mathbb{R}^p) = \blupp(U \cap U',(U \cap U') \cap \mathbb{R}^p)\]

as subsets of $\blupp(\mathbb{R}^n, \mathbb{R}^p)$. This fact is readily verified by tracing the definitions.

Now, if $(\base,\subbase)$ is a pair of smooth manifolds, let $\{(U_\alpha,\varphi_\alpha)\}_{\alpha \in I}$ be an adapted atlas (to $\subbase$) on $\base$ such that whenever $\subbase \cap U_\alpha \neq \emptyset$, we have $\varphi_\alpha(V_\alpha) = \varphi_\alpha(U_\alpha) \cap \mathbb{R}^p$ (where $V_\alpha \coloneqq U_\alpha \cap \subbase$). Let $\alpha,\beta \in I$ and write $U_{\alpha\beta}$ (and $V_{\alpha\beta}$) for the open set $U_\alpha \cap U_\beta$ (resp. $V_{\alpha} \cap V_{\beta}$), and $\varphi_{\alpha\beta}$ for the transition map $\varphi_\alpha \circ \varphi_\beta^{-1}$. From Corollary <ref> we obtain the map

\[\blupp(\varphi_\beta(U_{\alpha\beta}), \varphi_\beta(U_{\alpha\beta} \cap \subbase)) \xra{\blupp(\varphi_{\alpha\beta})} \blupp(\varphi_\alpha(U_{\alpha\beta}), \varphi_\alpha(U_{\alpha\beta} \cap \subbase)).\]

\[\blupp^{\sim}(\base,\subbase) = \bigsqcup_{\alpha \in I} \blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))\]

and define an equivalence relation $\sim$ on it such that

\begin{align*}
z &\sim w \textnormal{ (where } z \in \blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha)) \textnormal{ and } w \in \blupp(\varphi_\beta(U_\beta),\varphi_\beta(V_\beta))\textnormal{)} \textnormal{ if and only if } \\
z &\in \blupp(\varphi_\alpha(U_{\alpha\beta}),\varphi_\alpha(V_{\alpha\beta})), w \in \blupp(\varphi_\beta(U_{\alpha\beta}),\varphi_\beta(V_{\alpha\beta})), \textnormal{ and } \blupp(\varphi_{\alpha\beta})(w)=z.
\end{align*}

By Corollary <ref> and Remark <ref>, this is indeed an equivalence relation and the resulting quotient space

\[eq: \blupp^{\sim}(\base,\subbase) \ra \blupp(\base,\subbase)\]

is (set-theoretically) the blow-up of $\subbase$ in $\base$. We define a smooth structure on it by requiring that $eq|_{\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))}$ is a diffeomorphism onto its image for all $\alpha \in I$.
The smooth structure on $\blupp(\base,\subbase)$ is well-defined.

First of all, the maps $eq|_{\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))}$ are injective; this is obvious by definition of the equivalence relation $\sim$. It follows that the equivalence relation is open. Indeed,
to prove that $eq$ is an open map, we may restrict to the open sets $\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))$, but in this case it is obvious. The maps $eq|_{\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))}$ are now seen to be open (continuous) embeddings into $\blupp(\base,\subbase)$. The smooth structure on $\blupp(\base,\subbase)$ is defined by taking as open cover the subsets $eq(\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha)))$ together with the maps $eq|_{\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))}^{-1}$ post-composed with a chart on $\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha))$. Observe that the charts will be compatible using Remark <ref> and Corollary <ref>. This proves the statement.

Observe that if $\base$ is Hausdorff, then $\blupp(\base,\subbase)$ will be Hausdorff. To show this, observe that it suffices to prove that if $z,w \in \blupp(\base,\subbase)$, then we must have $z,w \in eq(\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha)))$ for some $\alpha \in I$. Indeed, if, say, $z \in eq(\blupp(\varphi_\beta(U_\beta),\varphi_\beta(V_\beta)))$ and $w \in eq(\blupp(\varphi_\gamma(U_\gamma),\varphi_\gamma(U_\gamma \cap \subbase)))$, then, since $eq^{-1}$ is well-defined when restricted to any of these open sets, we can consider the map

\[eq(\blupp(\varphi_\beta(U_\beta),\varphi_\beta(V_\beta))) \xra{eq^{-1}} \blupp(\varphi_\beta(U_\beta),\varphi_\beta(V_\beta)) \xra{\bldownn} \varphi_\beta(U_\beta) \xra{\varphi_\beta^{-1}} U_\beta \hookrightarrow \base\]

and similarly the map $eq(\blupp(\varphi_\gamma(U_\gamma),\varphi_\gamma(U_\gamma \cap \subbase))) \ra \base$. If the image of $z$ under the first map equals the image of $w$ under the second map, then there exists a chart $(U_\alpha,\varphi_\alpha)$ containing the common image. In this case it is then obvious that $z,w \in eq(\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha)))$. If the images of $z$ and $w$ under the respective maps are different, then we can find $\alpha',\beta' \in I$ such that $z \in U_{\alpha'}$, $w \in U_{\beta'}$, and $U_{\alpha'} \cap U_{\beta'} = \emptyset$. Then, we may assume $\varphi_{\alpha'}(U_{\alpha'}) \cap \varphi_{\beta'}(U_\beta') = \emptyset$, so we can consider the chart $(U_{\alpha'} \cup U_{\beta'},\varphi_{\alpha'} \cup \varphi_{\beta'}) \eqqcolon (U_{\alpha},\varphi_{\alpha})$ (where $\varphi_{\alpha'} \cup \varphi_{\beta'}: U_{\alpha'} \cup U_{\beta'} \ra \varphi_{\alpha'}(U_{\alpha'}) \cup \varphi_{\beta'}(U_\beta')$ is the map that sends $a$ to $\varphi_{\alpha'}(a)$ and $b \in U_{\beta'}$ to $\varphi_{\beta'}(b)$). It is clear now that $z,w \in eq(\blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha)))$. This shows that indeed $\blupp(\base,\subbase)$ is Hausdorff if $\base$ is Hausdorff.

We will now prove that this blow-up construction really is equivalent to our earlier blow-up construction. To do this, recall that the blow-up has a universal property (see Proposition <ref>). So, we only have to prove that the universal property holds for our new blow-up construction. This will follow from the following lemma (compare with Lemma <ref>).
Whenever $f: (\base,\subbase) \ra (\basetwo,\subbasetwo)$ is a map of pairs such that (1) $\subbasetwo \subset \basetwo$ is of codimension $1$, (2) $f^{-1}(\subbasetwo)=\subbase$, and (3) $d_\normal f$ is fiberwise injective, then there is a unique smooth map $\blupp(f): \blupp(\base,\subbase) \ra \blupp(\basetwo,\subbasetwo)$ such that the diagram

(,) r(f) d     (,) d

Let $x \in \base$ and let $(U,\varphi)$ and $(V,\psi)$ be adapted charts around $x \in \base$ and $f(x) \in \basetwo$ respectively. Then

\[(\psi \circ f \circ \varphi^{-1})^{-1}(\psi(V \cap \subbasetwo)) = \varphi(U \cap f^{-1}(V \cap \subbasetwo)) = \varphi(U \cap f^{-1}(V) \cap \subbase)\]
by the assumption that $f^{-1}(\subbasetwo)=\subbase$. Therefore, by Corollary <ref>, we obtain the smooth map $\blupp(\psi \circ f \circ \varphi^{-1})$ such that the diagram

(φ(U ∩f^-1(V)), φ(U ∩f^-1(V) ∩)) r(ψ∘f ∘φ^-1) d     (ψ(V), ψ(V ∩)) d

φ(U ∩f^-1(V)) rψ∘f ∘φ^-1     ψ(V)

commutes. Since $\blupp(\psi \circ f \circ \varphi^{-1}) = \blupp(\psi) \circ \blupp(f) \circ \blupp(\varphi)^{-1}$ by Corollary <ref>, we see that $\blupp(f)$ is a smooth map which has the desired property.

We will discuss here the universal property of $\blupp(\base,\subbase)$ that are similar to $\blup(\base,\subbase)$. First of all, using notation as in the discussion after Remark <ref>, notice that the blow-down maps

\[(\bldownn)_\alpha: \blupp(\varphi_\alpha(U_\alpha),\varphi_\alpha(V_\alpha)) \ra \varphi_\alpha(U_\alpha)\]

assemble into a smooth map $\bldownn: \blupp(\base,\subbase) \ra \base$. The restriction of $\blupp^\sim(\base,\subbase)$ to

\[\bigsqcup_{\alpha \in I} (\bldownn)_\alpha^{-1}(\varphi_\alpha(V_\alpha))\]

gives rise to a closed embedded submanifold $\mathbb{P}_{\textnormal{pc}}(\base,\subbase) \subset \blupp(\base,\subbase)$ which is of codimension $1$, and one can easily check that $\bldownn^{-1}(\subbase)=\mathbb{P}_{\textnormal{pc}}(\base,\subbase)$. We also have the following property: if $\subbase$ is of codimension $1$, then $\blupp(\base,\subbase)$ can be canonically identified with $\base$. Indeed, this follows simply by the fact that $\mathbb{R}^p \times S^{q-1} \times \mathbb{R} = \mathbb{R}^p \times \{\pm 1\} \times \mathbb{R}$. Now, like in Proposition <ref>, one can check (locally) that $d_\normal \bldownn$ is fiberwise injective. From Lemma <ref>, and using the same argument as in Proposition <ref>, we see that $(\blupp(\base,\subbase),\mathbb{P}_{\textnormal{pc}}(\base,\subbase))$ satisfies the same universal property as $(\blup(\base,\subbase),\mathbb{P}(\base,\subbase))$.

From the former remark, we see that $(\blupp(\base,\subbase),\mathbb{P}_{\textnormal{pc}}(\base,\subbase))$ and $(\blup(\base,\subbase),\mathbb{P}(\base,\subbase))$ are isomorphic as pairs of manifolds by a unique isomorphism. More explicitly, we have:

The functors $\blup$ and $\blupp$ (from the category of pairs of manifolds, together with maps of pairs $f: (\base,\subbase) \ra (\basetwo,\subbasetwo)$ for which $f^{-1}(\subbasetwo)=\subbase$ and $d_\normal f$ is fiberwise injective, to the category of manifolds) are inverse to each other.

Consider the morphism of functors $\mathfrak{B}: \blup \ra \blupp$ given by

\begin{align*}
\mathfrak{B}(\base,\subbase)&: \blup(\base,\subbase) \ra \blupp(\base,\subbase); \\
z &\mapsto
\begin{cases}
[\varphi^1(y),\frac{d_\normal \varphi(y)\xi}{\Vert d_\normal \varphi(y)\xi \Vert},0] & \text{if}\ z=[y,\xi] \\
[\varphi^1(x),\frac{t}{\vert t \vert}\frac{\varphi^2(x)}{\Vert \varphi^2(x) \Vert},\frac{t}{\vert t \vert}\Vert \varphi^2(x) \Vert] & \text{if}\ z=[x,t]
\end{cases}
\end{align*}

where $(U,\varphi)$ is an adapted chart around $\bldown(z)$. To see that this map is well defined, suppose $(V,\psi)$ is another adapted chart around $\bldown(z)$. If $z=[y,\xi]$, then

\begin{align*}
[\varphi^1(y),\frac{d_\normal \varphi(y)\xi}{\Vert d_\normal \varphi(y)\xi \Vert}] &= [\psi^1 \circ \varphi^{-1}(\varphi^1(y)),\frac{d_\normal(\psi \circ \varphi^{-1})(\varphi^1(y))}{\Vert d_\normal(\psi \circ \varphi^{-1})(\varphi^1(y)) \Vert} \circ \frac{d_\normal \varphi(y)\xi}{\Vert d_\normal \varphi(y)\xi \Vert}]  \\
&= [\psi^1(y),\frac{d_\normal \psi(y)\xi}{\Vert d_\normal \psi(y)\xi \Vert}]
\end{align*}

and if $z=[x,t]$, then, similarly,

\begin{align*}
[\varphi^1(x),\frac{t}{\vert t \vert}\frac{\varphi^2(x)}{\Vert \varphi^2(x) \Vert},\frac{t}{\vert t \vert}\Vert \varphi^2(x) \Vert] %&= [\psi^{1} \circ \varphi^{-1}(\varphi^2(y)),\frac{\psi^2 \circ \varphi^{-1}(\varphi^2(y))}{\Vert \psi^2 \circ \varphi^{-1}(\varphi^2(y)) \Vert}\frac{t}{\vert t \vert}\frac{\varphi^2(y)}{\Vert \varphi^2(y) \Vert},\Vert \psi^{2} \circ \varphi^{-1}(\varphi^2(y))\Vert\frac{t}{\vert t \vert}\Vert \varphi^2(y) \Vert] \\
&= [\psi^1(x),\frac{t}{\vert t \vert}\frac{\psi^2(x)}{\Vert \psi^2(x) \Vert},\frac{t}{\vert t \vert}\Vert \psi^2(x) \Vert].
\end{align*}

It remains to show that the map $\mathfrak{B}(\base,\subbase)$ is smooth for all pairs $(\base,\subbase)$, and that the diagrams

(,) r(f) d𝔅(,)     (,) d𝔅(',')

(,) r(f)     (,)

commute for all maps of pairs $f: (\base,\subbase) \ra (\basetwo,\subbasetwo)$ for which $f^{-1}(\subbasetwo)=\subbase$ and $d_\normal f$ is fiberwise injective. All of these statements follow from the observation that the map $\mathfrak{B}(\base,\subbase)$ is obtained from Lemma <ref> applied to the map of pairs

\[\bldown: (\blup(\base,\subbase),\mathbb{P}(\base,\subbase)) \ra (\base,\subbase).\]

This morphism of functors has an inverse $\mathfrak{B}^{-1}: \blupp \ra \blup$ given by

\begin{align*}
\mathfrak{B}(\base,\subbase)&: \blupp(\base,\subbase) \ra \blup(\base,\subbase); \\
z=[x,\theta,t] &\mapsto
\begin{cases}
[\varphi^{-1}(x),d_\normal\varphi^{-1}(x)\theta,0] & \text{if}\ t=0 \\
[\varphi^{-1}(x,t \cdot \theta),t] & \text{if}\ t\neq0.
\end{cases}
\end{align*}

where $(U,\varphi)$ is an adapted chart around $\bldownn(z)$. Using the same argument, but now applying Lemma <ref> to $\bldownn$, the result follows.

We end this section with two remarks on other variations of blow-ups of manifolds.

Another way to approach blow-ups, is by mimicking the algebraic blow-up construction explained in Section <ref>: since we know that blow-ups have to satisfy

\[\blup(\base \times \basetwo, \subbase \times \basetwo) \cong \blup(\base,\subbase) \times \basetwo,\]

we only have to describe, for the local description, the case

\[\subbase \coloneqq \{0\} \subset \mathbb{R}^n \eqqcolon \base.\]

To do this, observe that

\[\blup_{\textnormal{ag}}(\mathbb{R}^n,0) \coloneqq \{((x_1,\dots,x_n),[y_1:\cdots:y_n]) \mid x_iy_j=y_ix_j \textnormal{ for all } 1 \le i,j \le n\} \subset \mathbb{R}^n \times \mathbb{RP}^{n-1}\]

(the subscript “ag” refers to algebraic geometry) is a closed embedded submanifold of codimension $n-1$ (see Proposition <ref>).
We now define $\blup_{\textnormal{ag}}(\mathbb{R}^n,\mathbb{R}^p) \coloneqq \blup_{\textnormal{ag}}(\mathbb{R}^q,0) \times \mathbb{R}^p$. That is, $\blup_{\textnormal{ag}}(\mathbb{R}^n,\mathbb{R}^p)$ is equal to

\[\{((x_1,\dots,x_n),(y_{p+1}:\cdots:y_n)) \mid x_iy_j=y_ix_j \textnormal{ for all } p+1 \le i,j \le n\} \subset \mathbb{R}^n \times \mathbb{RP}^{n-p-1}.\]

Now observe that

\[S^{n-1} \times \mathbb{R} \ra \blup_{\textnormal{ag}}(\mathbb{R}^n,0) \textnormal{ given by } (\theta,t) \mapsto (t\theta,\theta)\]

(where we view $t\theta \in \mathbb{R}^n$ and $\theta \in \mathbb{RP}^{n-1}$) is $\mathbb{Z}_2$-equivariant, so it descends to a smooth map

\[\blupp(\mathbb{R}^n,0) \ra \blup_{\textnormal{ag}}(\mathbb{R}^n,0),\]

which is readily verified to be a diffeomorphism. So, by tracing the steps we made in this section, and using the above diffeomorphism repeatedly, we can proceed to prove the analogues of the statements for “$\blup_{\textnormal{ag}}$” to obtain another way to approach blow-ups for general pairs of manifolds $(\base,\subbase)$.

In view of Remark <ref>, there is yet another way to perform the blow-up construction. Recall that we can glue manifolds $M_1$ and $M_2$ along a diffeomorphism $\varphi: U_1 \ra U_2$, defined between open subsets $U_1 \subset M_1$ and $U_2 \subset M_2$ to a new manifold $M_1 \cup_{\varphi} M_2$. Using a tubular neighbourhood $\chi: (\normal(\base,\subbase),0_\subbase) \ra (\base,\subbase)$, we can describe the blow-up using this gluing construction (compare with Remark <ref>). In fact, what we will do is mimic the constructions explained in this section, and apply them to the normal bundle instead of locally to $\mathbb{R}^n$. First observe that

\[\widetilde{\normal}(\base,\subbase) \coloneqq \{(y,\xi_1,[y,\xi_2]) \in \normal(\base,\subbase) \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase) \mid \xi_1=\lambda \xi_2 \textnormal{ for some } \lambda \in \mathbb{R}^\times\}\]

(where $\pi_\subbase^\normal: \normal(\base,\subbase) \ra \subbase$ and $\pi_\subbase^{\mathbb{P}}: \mathbb{P}(\base,\subbase) \ra \subbase$ are the projection maps)
is a smooth manifold (note: locally, it is precisely given by the blow-up “$\blup_{\textnormal{ag}}(\base,\subbase)$” from Remark <ref>). Alternatively, we can choose a metric on $\base$ and realise the sphere bundle

\[S(\base,\subbase) \coloneqq S(\normal(\base,\subbase)) = (\normal(\base,\subbase) \setminus 0_\subbase)/\mathbb{R}_{>0},\]

where $\mathbb{R}_{>0}$ acts on $\normal(\base,\subbase) \setminus 0_\subbase$ by scalar multiplication (this construction works for more general vector bundles; see Lemma <ref>), as $\{(y,\xi) \in \normal(\base,\subbase) \mid \Vert\xi\Vert=1\}$. We have a free (and proper) $\mathbb{Z}_2$-action on $S(\base,\subbase) \times \mathbb{R}$ given by

\[\tau \cdot (y,\xi,t) = (y,-\xi,-t).\]

Then the map

\[S(\base,\subbase) \times \mathbb{R} \ra \normal(\base,\subbase) \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase) \textnormal{ given by } (y,\xi,t) \mapsto (y,t \cdot \xi, [y,\xi])\]

is smooth and intertwines the $\mathbb{Z}_2$-action (with trivial action on $\normal(\base,\subbase) \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase)$), so it descends to a smooth map

\[(S(\base,\subbase) \times \mathbb{R})/\mathbb{Z}_2 \ra \normal(\base,\subbase) \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase).\]

This map is an embedding, and it maps onto $\widetilde{\normal}(\base,\subbase)$ (note: locally, it is precisely given by the blow-up “$\blupp(\base,\subbase)$” explained in this section). Now, $\widetilde{\normal}(\base,\subbase)$ comes with a smooth map

\[\widetilde{\normal}(\base,\subbase) \xra{\pr_1} \normal(\base,\subbase)\]

which is a diffeomorphism onto $\normal(\base,\subbase) \setminus 0_\subbase$ when restricted to $\widetilde{\normal}(\base,\subbase) \setminus (0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase))$. Therefore, we can glue $\widetilde{\normal}(\base,\subbase)$ and $\base \setminus \subbase$ via the map

\[\widetilde\chi \coloneqq \chi \circ \pr_1: \widetilde{\normal}(\base,\subbase) \setminus (0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase)) \ra \base \setminus \subbase\]

(which is a diffeomorphism onto an open subset of $\base \setminus \subbase$) to the smooth manifold

\[\blup_{\textnormal{tub}}(\base,\subbase) \coloneqq \widetilde\normal(\base,\subbase) \cup_{\widetilde\chi} \base \setminus \subbase.\]

We claim that this manifold is diffeomorphic to $\blup(\base,\subbase)$.To see this, notice that the blow-down map for $\blup_{\textnormal{tub}}(\base,\subbase)$ is given by

\[\bldown_{\textnormal{tub}} \coloneqq \chi \circ \pr_1 \cup_{\widetilde\chi} \iota: \blup_{\textnormal{tub}}(\base,\subbase) \ra \base,\]

where $\iota: \base \setminus \subbase \hookrightarrow \base$ is the inclusion map (note: $\chi \circ \pr_1$ has domain $\widetilde\normal(\base,\subbase)$).
Notice now that $\widetilde\normal(\base,\subbase)$ contains the codimension $1$ submanifold $0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase)$, that $\bldown_{\textnormal{tub}}^{-1}(\subbase) = 0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase)$, and that $d_\normal\bldown_{\textnormal{tub}}$ is fiberwise injective (since $\chi \circ \pr_1$ is), so we obtain a smooth map

\[\blup(\bldown_{\textnormal{tub}}): \blup(\blup_{\textnormal{tub}}(\base,\subbase),0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase)) \cong \blup_{\textnormal{tub}}(\base,\subbase) \ra \blup(\base,\subbase).\]

Notice that it restricts to the canonical diffeomorphism $0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase) \cong \mathbb{P}(\base,\subbase)$. Moreover, the inclusion

\[\base \setminus \subbase \hookrightarrow \blup_{\textnormal{tub}}(\base,\subbase)\]

extends uniquely to a smooth map $\blup(\base,\subbase) \ra \blup_{\textnormal{tub}}(\base,\subbase)$ that restricts to the canonical diffeomorphism $\mathbb{P}(\base,\subbase) \cong 0_\subbase \tensor[_{\pi_\subbase^\normal}]{\times}{_{\pi_\subbase^{\mathbb{P}}}} \mathbb{P}(\base,\subbase)$. This shows that $\blup(\bldown_{\textnormal{tub}})$ is a diffeomorphism.

§.§.§ Blow-ups of groupoids and algebroids: a different approach

In <cit.>, a different blow-up construction is explained for Lie groupoids. There, they are only interested in blowing up Lie groupoids for which the base manifold does not change: this happens if and only if the submanifold of the base manifold is of codimension $1$. We will phrase the differences of our earlier blow-up construction for Lie groupoids (from <cit.>), and the one from <cit.> by identifying two main differences. First of all, the blow-up construction used there, on the level of manifolds, is the construction from Section <ref> (see also Remark <ref>). This is one of the main differences. For the other main difference, note that the key to defining blow-ups for both Lie groupoids and Lie algebroids, lies in Remark <ref>: it says that if $f: (\base, \subbase) \ra (\basetwo,\subbasetwo)$ is a smooth map of pairs, then we obtain canonically a smooth map $\blup(f): \blup_f(\base,\subbase) \ra \blup(\basetwo,\subbasetwo)$, where

\[\blup_f(\base,\subbase) \coloneqq (\DNC(\base,\subbase) \setminus \DNC(f)^{-1}(\subbasetwo \times \mathbb{R}))/\mathbb{R}^\times.\]

If we find the analogue of this Remark in the setting of Section <ref> (that is, by not having to factor through the deformation to the normal cone), then one can, again, define the structure maps this way, and proceed similarly as in the proofs of Theorem <ref>, Theorem <ref>, and Theorem <ref>. Notice that in <cit.>, there is no geometric construction for the blow-up of Lie algebroids. They do, however, give the description on the level of sections (in the codimension $1$ case; it is called the lower elementary modification). For a pair of Lie algebroids $(\algebroid,\subalgebroid)$, this space of sections is $\Gamma(\algebr,\subalgebr)$, and we have seen in Theorem <ref>, that, indeed, our Lie algebroid construction has $\Gamma(\algebr,\subalgebr)$ as its space of sections (in the codimension $1$ case).

We will now prove the analogue of Remark <ref> in the setting of Section <ref>. We fix here a smooth map of pairs $f: (\base, \subbase) \ra (\basetwo,\subbasetwo)$.

We have a canonical smooth map

\begin{align*}
\blupp(f): \blup_{\textnormal{pc}, f}(\base,\subbase) \ra \blupp(\basetwo,\subbasetwo)
\end{align*}

\[\blup_{\textnormal{pc}, f} \coloneqq \blupp(\base,\subbase) \setminus \overline{\bldownn^{-1}(f^{-1}(\subbasetwo) \setminus \subbase)}.\]

Moreover, under the canonical diffeomorphism $\blupp(\base,\subbase) \cong \blup(\base,\subbase)$ from Proposition <ref>, we have $\blup_{\textnormal{pc}, f} \cong \blup_f(\base,\subbase)$.

Since we already proved the analogue of this statement for our earlier blow-up construction, it suffices to prove the second statement. We denote $Z \coloneqq \bldown^{-1}(f^{-1}(\subbasetwo) \setminus \subbase) \subset \blup(\base,\subbase)$. The statement follows if we prove that

\[\blup_f(\base,\subbase) = \blup(\base,\subbase) \setminus \overline{Z}.\]

Since $\bldown^{-1}(\subbase) = \mathbb{P}(\base,\subbase)$, and $f^{-1}(\subbasetwo) \subset \base$ is closed, we see that $(\base \setminus \subbase) \cap \overline{Z} = f^{-1}(\subbasetwo) \setminus \subbase$. Therefore, we only have to prove the following claim:

\begin{equation}\label{eq: claim in proper transform}
\textnormal{ if } z \coloneqq [y,\xi] \in \mathbb{P}(\base,\subbase), \textnormal{ then } z \textnormal{ is in the closure of } Z \textnormal{ if and only if } d_\normal f(y)\xi = 0.
\end{equation}

We fix an element $z \coloneqq [y,\xi] \in \mathbb{P}(\base,\subbase)$, and denote $q_\DNC: P \rightarrowdbl \blup(\base,\subbase)$ for the quotient projection, where $P \coloneqq \DNC(\base,\subbase) \setminus (\subbase \times \mathbb{R})$.

First, we will show that, if $d_\normal f(y)\xi \neq 0$, then $z$ is not in the closure of $Z$. Notice that $z$ is in the closure of $Z$ if and only if every $\mathbb{R}^\times$-invariant open subset $U \ni (y,\xi)$ of $P$ intersects

\[Z_\DNC \coloneqq q_\DNC^{-1}(Z) = (f^{-1}(\subbasetwo) \setminus \subbase) \times \mathbb{R}^\times = f^{-1}(\subbasetwo) \times \mathbb{R}^\times \setminus (\subbase \times \mathbb{R}^\times) \subset P.\]

We will show that if $d_\normal f(y)\xi \neq 0$, then we can find an open invariant subset of $P$ that is disjoint from $Z_\DNC$. Indeed, $\DNC(f)$ is continuous, so we can find for all invariant open subsets $V$ around $\DNC(f)(y,\xi)$ an open subset $U \subset P$ around $(y,\xi)$ such that $\DNC(f)(U) \subset V$ (we use here that $\subbase \times \mathbb{R} \subset \DNC(\base,\subbase)$ is closed). By equivariance of $\DNC(f)$ and $V$, $\DNC(f)(\textstyle\cup_{\lambda \in \mathbb{R}^\times} \lambda \cdot U) \subset V$, so we can assume $U$ is invariant as well. By assumption, $\DNC(f)(y,\xi)=d_\normal f(y)\xi \not\in \subbasetwo \times \mathbb{R}$, so we may assume $V$ is disjoint from $\subbasetwo \times \mathbb{R}$ (because $\subbasetwo \times \mathbb{R}$ is closed in $\DNC(\basetwo,\subbasetwo)$). But now $\DNC(f)(U)$ is disjoint from $\subbasetwo \times \mathbb{R}$, so $U$ is disjoint from $Z_\DNC$.

Conversely, suppose $d_\normal f(y)\xi=0$. If $z$ is not in the closure of $Z$, then there is an invariant open subset $W \ni (y,\xi)$ of $P \coloneqq \DNC(\base,\subbase) \setminus (\subbase \times \mathbb{R})$ that is disjoint from $Z_\DNC$. By shrinking $W$ if necessary, we may assume $W = \DNC(U,V) \cap P$ for some adapted chart $(U,\varphi)$ of $\base$ (with $V \coloneqq U \cap \subbase$). Now, since $d_\normal f(y)\xi = 0$, $\xi \in T_y\base$ spans a one-dimensional subspace $\langle \xi \rangle$ of $T_y\base$ which maps into $T_{f(y)}\subbasetwo$, but it has a trivial intersection with $T_y\subbase$. We can integrate $\langle \xi \rangle$ to an embedded $1$-dimensional submanifold $S$ of $U$. By construction, we can shrink $S$ such that $f$ maps $S$ into $\subbasetwo$, but this makes it clear that $S \times \mathbb{R}^\times \subset W$ has a non-trivial intersection with $Z_\DNC$. This contradicts our assumption on $W$. This proves (<ref>), so this concludes the proof.

Let $(\base,\subbase)$ be a pair of manifolds, and let $S \subset \base$. The closed set

\[\overoverline{S} \coloneqq \overline{\bldown^{-1}(S \setminus \subbase)} \subset \blup(\base,\subbase)\]

is called the proper transform of $S$.

With this notation, if $f: (\base,\subbase) \ra (\basetwo,\subbasetwo)$ is a map of pairs, then we have

\[\blup_f(\base,\subbase)=\blup(\base,\subbase) \setminus \overoverline{f^{-1}(\subbasetwo)}; \quad \blup_{f,g}(\base,\subbase)=\blup(\base,\subbase) \setminus (\overoverline{f^{-1}(\subbasetwo)} \cup \overoverline{g^{-1}(\subbasetwo)}).\]

In Definition <ref>, which is the analogue of Definition <ref> for varieties (in case of blowing up a point), we already hinted upon thinking about this as another way to blow-up:

Let $S \subset \base$ be a closed embedded submanifold such that $\subbase \subset S$. Then

\[\blup(S,\subbase) = \overoverline{S}\]

viewed as subspaces of $\blup(\base,\subbase)$ (see also Corollary <ref>).

Since $\bldown^{-1}(S \setminus \subbase) \subset \blup(S,\subbase)$, and $\blup(S,\subbase) \subset \blup(\base,\subbase)$ is closed (see Corollary <ref>), we see that $\overoverline{S} = \overline{\bldown^{-1}(S \setminus \subbase)} \subset \blup(S,\subbase)$. For the converse inclusion, observe that $\bldown^{-1}(S \setminus \subbase)$ is dense as a subspace of $\blup(S,\subbase)$, so if $[y,\xi] \in \mathbb{P}(S,\subbase) \subset \blup(S,\subbase)$, and $U \ni [y,\xi]$ is an open subset of $\blup(\base,\subbase)$, then $U \cap \blup(S,\subbase)$ intersects $\bldown^{-1}(S \setminus \subbase)$ non-trivially, which shows that $\blup(S,\subbase) \subset \overoverline{S}$. This proves the statement.

To illustrate the ideas presented in this section and the former section, let us repeat the simple example explained in Example <ref>.

We will reprove that the blow-up of $\{(0,0)\} \rra \{0\}$ in the pair groupoid $\mathbb{R} \times \mathbb{R} \rra \mathbb{R}$ yields $\mathbb{R}^\times \ltimes \mathbb{R}$. Indeed, recall that

\[\blupp(\mathbb{R}^2,0) = (S^1 \times \mathbb{R})/\mathbb{Z}_2,\]

\[\bldownn: \blupp(\mathbb{R}^2,0) \ra \mathbb{R}^2 \textnormal{ is given by } [\theta,t] \mapsto t\theta.\]

From this we see that

\begin{align*}
\bldownn^{-1}(\pr_2^{-1}(0) \setminus \{(0,0)\}) &= \{[\theta,t] \mid \theta=(\pm1,0), t\neq0\}; \\
\bldownn^{-1}(\pr_1^{-1}(0) \setminus \{(0,0)\}) &= \{[\theta,t] \mid \theta=(0,\pm1), t\neq0\},
\end{align*}

and, in turn, this shows that

\[\overoverline{\pr_2^{-1}(0)} = \{[\theta,t] \mid \theta=(\pm1,0)\}; \quad \overoverline{\pr_1^{-1}(0)} = \{[\theta,t] \mid \theta=(0,\pm1)\}.\]

\[\blup_{\textnormal{pc}, \source,\target}(\mathbb{R}^2,0) = ((S^1 \setminus \{\theta = (\theta_1,\theta_2) \in S^1 \mid \theta_1 \neq 0, \theta_2 \neq 0\}) \times \mathbb{R})/\mathbb{Z}_2.\]

Now, the smooth map

\[(S^1 \setminus \{\theta \in S^1 \mid \theta_1 \neq 0, \theta_2 \neq 0\}) \times \mathbb{R} \ra \mathbb{R}^\times \times \mathbb{R} \textnormal{ given by } (\theta,t) \mapsto (\frac{\theta_1}{\theta_2},t\theta_2)\]

is $\mathbb{Z}_2$-equivariant (with trivial action on $\mathbb{R}^\times \times \mathbb{R}$), so it descends to a smooth map

\begin{equation}\label{eq: diff to R^times times R}
\blup_{\textnormal{pc},\source,\target}(\mathbb{R}^2,0) \ra \mathbb{R}^\times \times \mathbb{R},
\end{equation}

and this map has a smooth inverse, namely $(\lambda,t) \mapsto [\tfrac{1}{\sqrt{1+\lambda^{-2}}}, \tfrac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}, t\tfrac{\lambda}{\sqrt{1+\lambda^{-2}}}+t\tfrac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}]$; here is a direct computation:

\begin{align*}
[\frac{1}{\sqrt{1+\lambda^{-2}}}&, \frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}, t\frac{\lambda}{\sqrt{1+\lambda^{-2}}}+t\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}] \\
&\mapsto (\frac{\frac{1}{\sqrt{1+\lambda^{-2}}}}{\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}},(t\frac{\lambda}{\sqrt{1+\lambda^{-2}}}+t\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}) \cdot \frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}) = (\lambda,t), \textnormal{ and } \\
&\mapsto [\frac{1}{\sqrt{1+\frac{\theta_2^2}{\theta_1^2}}}, \frac{\frac{\theta_2}{\theta_1}}{\sqrt{1+\frac{\theta_2^2}{\theta_1^2}}}, t\theta_2\frac{\frac{\theta_1}{\theta_2}}{\sqrt{1+\frac{\theta_2^2}{\theta_1^2}}}+t\theta_2\frac{\frac{\theta_2}{\theta_1}}{\sqrt{1+\frac{\theta_2^{2}}{\theta_1^2}}}] \\
&= [\frac{\textnormal{sign}(\theta_1)\theta_1}{\sqrt{\theta_1^2+\theta_2^2}}, \frac{\textnormal{sign}(\theta_1)\theta_2}{\sqrt{\theta_1^2+\theta_2^2}}, t\frac{\textnormal{sign}(\theta_1)\theta_1^2}{\sqrt{\theta_1^2+\theta_2^2}}+t\frac{\textnormal{sign}(\theta_1)\theta_2^2}{\sqrt{\theta_1^2+\theta_2^2}}] = [\theta,t].
\end{align*}

To see that the diffeomorphism (<ref>) is a Lie groupoid isomorphism, we will only check
that, under this diffeomorphism, $\blup(\source)$ and $\blup(\target)$ become the source map and the target map of $\mathbb{R}^\times \ltimes \mathbb{R}$, respectively (that this works for the other structure maps as well is readily verified by doing similar computations). Indeed,

\[\blup(\source)([\theta,t])=t\theta_2; \quad \blup(\target)([\theta,t])=t\theta_1\]
(see Lemma <ref>), and, indeed,

\[\source_{\mathbb{R}^\times \ltimes \mathbb{R}}(\frac{\theta_1}{\theta_2},t\theta_2)=t\theta_2; \quad \target_{\mathbb{R}^\times \ltimes \mathbb{R}}(\frac{\theta_1}{\theta_2},t\theta_2) = t\theta_1,\]

\begin{align*}
\blup(\source)([\frac{1}{\sqrt{1+\lambda^{-2}}}&, \frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}, t\frac{\lambda}{\sqrt{1+\lambda^{-2}}}+t\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}]) \\
&= (t\frac{1}{\sqrt{1+\lambda^{-2}}}+t\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}) \cdot \frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}} = t \\
\blup(\target)([\frac{\lambda}{\sqrt{1+\lambda^{-2}}}&, \frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}, t\frac{\lambda}{\sqrt{1+\lambda^{-2}}}+t\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}]) \\
&= (t\frac{\lambda}{\sqrt{1+\lambda^{-2}}}+t\frac{\lambda^{-1}}{\sqrt{1+\lambda^{-2}}}) \cdot \frac{1}{\sqrt{1+\lambda^2}} = \lambda t.
\end{align*}

From this we see that $\blup_{\textnormal{pc},\source,\target}(\mathbb{R}^2,0) \cong \mathbb{R}^\times \ltimes \mathbb{R}$, as expected.

§.§ Examples of blow-ups of pairs of Lie groupoids and Lie algebroids

We now turn to examples of blow-ups of Lie groupoids and Lie algebroids. The first type of examples comes from restricting to a particular class of submanifolds of the base manifold, called saturated submanifolds.

§.§.§ Blow-up along a saturated submanifold

We will start this section with discussing the blow-up of an action groupoid. Immediately after, we will discuss the blow-up of an action algebroid. After that, we will use these examples to show that a large class of blow-up groupoids and algebroids can be described, canonically, as action groupoids/action algebroids. As already mentioned, we will focus on the following specific class of submanifolds that we blow-up in the base manifold.

If $\groupoid$ is a groupoid, we call a submanifold $\subbase \subset \base$ for which

\[\source^{-1}(\subbase)=\target^{-1}(\subbase)\]

(i.e. $\subbase$ is a union of orbits) a saturated submanifold of $\base$ (with respect to $\group$). Similarly, if $\algebroid$ is an algebroid, we call a submanifold $\subbase \subset \base$ which is a union of orbits a saturated submanifold of $\base$ (with respect to $\algebr$).

Let $(\basetwo,\subbasetwo)$ be a pair of manifolds such that $\basetwo$ is equipped with a $\groupoid$ action with moment map $\mu: \basetwo \ra \base$, and assume that $\subbasetwo$ is saturated with respect to this action. Then

\[\blup_{\source,\target}(\group \ltimes \basetwo, \group \ltimes \subbasetwo) = \blup(\group \ltimes \basetwo, \group \ltimes \subbasetwo),\]

and we can describe a $\group$-action on $\blup(\basetwo,\subbasetwo)$, with moment map $\mu \circ \bldown: \blup(\basetwo,\subbasetwo) \ra \base$, such that

\[\blup(\group \ltimes \basetwo, \group \ltimes \subbasetwo) \cong \group \ltimes \blup(\basetwo,\subbasetwo)\]

as Lie groupoids.

The first statement follows from the fact that $\subbasetwo \subset \basetwo$ is saturated, and that $\source_{\group \ltimes \basetwo} = \pr_2$, so that $d_\normal\source_{\group \ltimes \basetwo}$ is fiberwise a bijection, and therefore so is $d_\normal\target_{\group \ltimes \basetwo}$ (by the relation $d_\normal\source_{\group \ltimes \basetwo}=d_\normal\target_{\group \ltimes \basetwo} \circ d_\normal\iota_{\group \ltimes \basetwo}$). Now, the diffeomorphism

\begin{equation}\label{eq: blup(G times M,G times N) = G times blup(M,N)}
\blup(\group \times \basetwo, \group \times \subbasetwo) \xra{\sim} \group \times \blup(\basetwo,\subbasetwo) \textnormal{ given by } z \mapsto
\begin{cases}
(g,[y,\xi])    & \text{if}\ z=[g,y,0,\xi] \\
(g,[x,1])    & \text{if}\ z=[g,x,1]
\end{cases}
\end{equation}

from Proposition <ref>, restricts to a diffeomorphism

\begin{equation}\label{eq: blup(G times M,G times N) = G times blup(M,N) 2}
\blup(\group \tensor[_{\source}]{\times}{_{\mu}} \basetwo, \group \tensor[_{\source}]{\times}{_{\mu}} \subbasetwo) \xra{\sim} \group \tensor[_{\source}]{\times}{_{\mu \circ \bldown}} \blup(\basetwo,\subbasetwo).
\end{equation}

Indeed, since $\blup(\group \tensor[_{\source}]{\times}{_{\mu}} \basetwo, \group \tensor[_{\source}]{\times}{_{\mu}} \subbasetwo) \subset \blup(\group \times \basetwo, \group \times \subbasetwo)$ and $\group \tensor[_{\source}]{\times}{_{\mu \circ \bldown}} \blup(\basetwo,\subbasetwo) \subset \group \times \blup(\basetwo,\subbasetwo)$ are embedded submanifolds, we only have to show that the restriction of (<ref>) is a bijection, but this is readily verified. Moreover, if we denote $\nu: \group \tensor[_{\source}]{\times}{_{\mu}} \basetwo \ra \basetwo$ for the $\group$-action on $\base$, then

\[\blup_\nu(\group \tensor[_{\source}]{\times}{_{\mu}} \basetwo,\group \tensor[_{\source}]{\times}{_{\mu}} \subbasetwo) = \blup(\group \tensor[_{\source}]{\times}{_{\mu}} \basetwo,\group \tensor[_{\source}]{\times}{_{\mu}} \subbasetwo),\]

because $d_\normal\target_{\group \ltimes \basetwo} = \nu$ is fiberwise a bijection.
Therefore, we can define a multiplication on $\blup(\basetwo,\subbasetwo)$ by setting

\[\nu_{\blup(\basetwo,\subbasetwo)} \coloneqq \blup(\nu): \group \tensor[_{\source}]{\times}{_{\mu \circ \bldown}} \blup(\basetwo,\subbasetwo) \cong \blup(\group \tensor[_{\source}]{\times}{_{\mu}} \basetwo,\group \tensor[_{\source}]{\times}{_{\mu}} \subbasetwo) \ra \blup(\basetwo,\subbasetwo).\]

It remains to show that the diffeomorphism (<ref>) is a morphism of Lie groupoids. This is, however, precisely how we defined the action on $\blup(\basetwo,\subbasetwo)$: $\blup(\target) = \blup(\nu)$ becomes $\target_{\group \ltimes \blup(\basetwo,\subbasetwo)} = \nu_{\blup(\basetwo,\subbasetwo)}$ under the diffeomorphism (<ref>) by construction. That the map intertwines the other structure maps as well, is readily verified. This proves the statement.

If a Lie group $G$ acts on $\base$, and $\subbase \subset \base$ is saturated with respect to the action, then

\[\blup_{\source,\target}(G \times \base, G \times \subbase) = \blup(G \times \base, G \times \subbase),\]

and we can describe a $G$-action on $\blup(\base,\subbase)$ such that

\[\blup(G \ltimes \base, G \ltimes \subbase) \cong G \ltimes \blup(\base,\subbase)\]

as Lie groupoids.

Let $(\basetwo,\subbasetwo)$ be a pair of manifolds such that $\basetwo$ is equipped with an $\algebroid$ action with moment map $\mu: \basetwo \ra \base$, and assume that $\subbasetwo$ is saturated with respect to this action. Then

\[\blup_\pi(\algebr \ltimes \basetwo, \algebr \ltimes \subbasetwo) = \blup(\algebr \ltimes \basetwo, \algebr \ltimes \subbasetwo),\]

and we can describe an $\algebr$-action on $\blup(\basetwo,\subbasetwo)$, with moment map $\mu \circ \bldown: \blup(\basetwo,\subbasetwo) \ra \base$, such that

\[\blup(\algebr \ltimes \basetwo, \algebr \ltimes \subbasetwo) \cong \algebr \ltimes \blup(\basetwo,\subbasetwo)\]

as Lie algebroids.

A lot of arguments are completely analogous to the proof of Proposition <ref>. For example, the first statement follows from the fact that $\subbasetwo \subset \basetwo$ is saturated, and that $\pi_{\algebr \ltimes \basetwo}=\pr_2$, so that $d_\normal\pi_{\algebr \ltimes \basetwo}$ is fiberwise a bijection. Also, we obtain a diffeomorphism

\[\varphi: \blup(\algebr \tensor[_{\pi}]{\times}{_{\mu}} \basetwo, \algebr \tensor[_{\pi}]{\times}{_{\mu}} \subbasetwo) \xra{\sim} \algebr \tensor[_{\pi}]{\times}{_{\mu \circ \bldown}} \blup(\basetwo,\subbasetwo)\]

in exactly the same way (by using Proposition <ref>). Now, that $\algebr$ acts on $\basetwo$, with moment map $\mu: M \ra \base$, can be equivalently described by saying that $\mu^*\algebr=\algebr \tensor[_{\pi}]{\times}{_{\mu}} \basetwo$ carries a Lie algebroid structure such that the map $\pr_1: \mu^*\algebr \ra \algebr$ is a morphism of Lie algebroids. Since the blow-down map

\[\bldown: \blup(\algebr \tensor[_{\pi}]{\times}{_{\mu}} \basetwo, \algebr \tensor[_{\pi}]{\times}{_{\mu}} \subbasetwo) \ra \algebr \tensor[_{\pi}]{\times}{_{\mu}} \basetwo\]

is a morphism of Lie algebroids, the map $\pr_1 \circ \bldown: \blup(\algebr \tensor[_{\pi}]{\times}{_{\mu}} \basetwo, \algebr \tensor[_{\pi}]{\times}{_{\mu}} \subbasetwo) \ra \algebr$ is too, and it is readily verified that this map is equal to $\pr_1 \circ \varphi$. This provides $\blup(M,N)$ with the desired $\algebr$-action, and $\varphi$ is an isomorphism of Lie algebroids by construction.

If a Lie algebra $\mathfrak{g}$ acts on $\base$, and $\subbase \subset \base$ is saturated with respect to the action, then

\[\blup_{\pi}(\mathfrak{g} \times \base, \mathfrak{g} \times \subbase) = \blup(\mathfrak{g} \times \base, \mathfrak{g} \times \subbase),\]

and we can describe a $\mathfrak{g}$-action on $\blup(\base,\subbase)$ such that

\[\blup(\mathfrak{g} \ltimes \base, \mathfrak{g} \ltimes \subbase) \cong \mathfrak{g} \ltimes \blup(\base,\subbase)\]

as Lie algebroids.

Notice that virtually the same proofs work if we replace “$\blup$” everywhere with “$\DNC$” in Propositions <ref> and <ref>.

In the results above we were in the situation that $\subbase \subset \base$ is saturated. By observing that $\group$ is canonically isomorphic to the action groupoid $\group \times \base$ (with moment map $\source$ and multiplication map $\target$), we obtain that the blow-up of a groupoid/algebroid along the restriction groupoid/algebroid with respect to a saturated submanifold results in an action groupoid/algebroid.

Let $\groupoid$ be a Lie groupoid. If $\subbase \subset \base$ is saturated, then we can describe a (canonical) $\group$-action on $\blup(\base,\subbase)$, with moment map $\bldown$, such that

\[(\bldown,\blup(\source)): \blup(\group,\group|_{\subbase}) \xra{\sim} \group \ltimes \blup(\base,\subbase)\]

is an isomorphism of Lie groupoids. Similarly, if $\algebroid$ is a Lie algebroid, and $\subbase \subset \base$ is saturated, then we can describe a (canonical) $\algebr$-action on $\blup(\base,\subbase)$, with moment map $\bldown$, such that

\[(\bldown,\blup(\pi)): \blup(\algebr,\algebr|_{\subbase}) \xra{\sim} \algebr \ltimes \blup(\base,\subbase)\]

is an isomorphism of Lie algebroids.

\begin{align*}
(\group \tensor[_{\source}]{\times}{_{\bldown}} \blup(\base,\subbase))|_{\mathbb{P}(\base,\subbase)} &\cong \group|_{\subbase} \ltimes \mathbb{P}(\base,\subbase) \cong \mathbb{P}(\group,\group|_{\subbase}) \textnormal{ and } \\
(\group \tensor[_{\source}]{\times}{_{\bldown}} \blup(\base,\subbase))|_{\base \setminus \subbase} &= \group|_{\base \setminus \subbase},
\end{align*}

we see that the action map $\nu: \group \tensor[_{\source}]{\times}{_{\bldown}} \blup(\base,\subbase) \ra \blup(\base,\subbase)$ is given by the map

\[g \cdot z \coloneqq
\begin{cases}
g \cdot [y,\xi]    & \text{if}\ z=[y,\xi] \\
[x,1]    & \text{if}\ z=[x,1]
\end{cases} =
\begin{cases}
[\target(g),d_\normal\target(g)\eta]    & \text{if}\ z=[\source(g),d_\normal\source(g)\eta] \\
[\target(g),1]    & \text{if}\ z=[\source(g),1]
\end{cases}\]

(note: if $z=[y,\xi] \in \mathbb{P}(\base,\subbase)$, then $(g,z) \in \group \tensor[_{\source}]{\times}{_{\bldown}} \blup(\base,\subbase)$ if and only if $g \in \group|_{\subbase}$

We also include the following two statements which might be instructive at this point.
Let $\groupoid$ be a Lie groupoid, and let $\subbase \subset \base$ be a saturated embedded submanifold.
Then there is a $\group|_{\subbase}$-action on $\normal(\base,\subbase)$, with moment map $\pi_\subbase: \normal(\base,\subbase) \ra \subbase$, such that the map

\[(\pi_\subbase^\group,d_\normal\source): \normal(\group,\group|_{\subbase}) \xra{\sim} \group|_{\subbase} \ltimes \normal(\base,\subbase)\]

is an isomorphism of Lie groupoids. Moreover, $\mathbb{P}_{\source,\target}(\group,\group|_{\subbase})=\mathbb{P}(\group,\group|_{\subbase})$, and the above map descends to an isomorphism of Lie groupoids $\mathbb{P}(\group,\group|_{\subbase}) \xra{\sim} \group|_{\subbase} \ltimes \mathbb{P}(\base,\subbase)$.

The $\group|_{\subbase}$-action on $\normal(\base,\subbase)$ is

\begin{align*}
\nu: \group|_{\subbase} \tensor[_{\source}]{\times}{_{\pi_\subbase}} \normal(\base,\subbase) &\ra \normal(\base,\subbase) \textnormal{ given by } (g,y,\xi) \mapsto (\target(g),d_\normal\target(g)\eta), \textnormal{ where } \\
\eta &\in \normal_g(\group,\group|_{\subbase}) \textnormal{ such that } \xi = d_\normal\source(g)\eta.
\end{align*}

To see that this is well-defined, suppose that also $\xi = d_\normal\source(g)\zeta$. Since $\source$ and $\target$ are submersions, $\source^{-1}(\subbase)=\target^{-1}(\subbase)$ is an embedded submanifold with

\[d\source(g)^{-1}(T_{\source(g)}\subbase) = T_g\source^{-1}(\subbase) = T_g\target^{-1}(\subbase) = d\target(g)^{-1}(T_{\target(g)}\subbase).\]

Since $d\source(g)(\eta-\zeta) \in T_{\source(g)}\subbase$, we see that $d\target(g)(\eta-\zeta) \in T_{\target(g)}\subbase$, so $d_\normal\target(g)\eta=d_\normal\target(g)\zeta$.

That the given map is a morphism of groupoids is now readily verified. To see that it is even an isomorphism, notice that

(,|_) [r] [d,shift left] [d,shift right]     |_ [d,shift left] [d,shift right]

(,) [r]

is a $\VB$-groupoid (see Section <ref>), and the given map is the map from Lemma <ref>, which is, in that lemma, proven to be a surjective submersion. Since the map is also injective in this case, it is a diffeomorphism, and therefore it is an isomorphism of Lie groupoids. The last statements follows, because the map restricts to an isomorphism

\[\normal_{\source,\target}(\group,\group|_{\subbase}) \xra{\sim} \group|_{\subbase} \ltimes (\normal(\base,\subbase) \setminus 0_\subbase)\]

as is readily verified, and it is obvious that the diffeomorphism $\mathbb{P}_{\source,\target}(\group,\group|_{\subbase}) \xra{\sim} \group|_{\subbase} \ltimes \mathbb{P}(\base,\subbase)$ it descends to is a morphism of Lie groupoids as well. Moreover, $\mathbb{P}_{\source,\target}(\group,\group|_{\subbase})=\mathbb{P}(\group,\group|_{\subbase})$
. This proves the statement.

By using that, for a Lie algebroid $\algebroid$ and $\subbase \subset \base$ a saturated embedded submanifold,

(,|_) [r] [d]     |_ [d]

(,) [r]

is a $\VB$-algebroid, and using Lemma <ref> instead, we see that a similar results holds for a Lie algebroid $\algebroid$ and a saturated submanifold $\subbase \subset \base$ (note: an action of $\algebr|_{\subbase}$, with moment map $\pi_\subbase$, is a Lie algebroid structure on $\pi_\subbase^*(\algebr|_{\subbase})$ such that the map $\pr_1: \pi_\subbase^*(\algebr|_{\subbase}) \ra \algebr|_{\subbase}$ is a morphism of Lie algebroids).

Let $\algebroid$ be a Lie algebroid, and let $\subbase \subset \base$ be a saturated embedded submanifold.
Then there is a $\algebr|_{\subbase}$-action on $\normal(\base,\subbase)$, with moment map $\pi_\subbase: \normal(\base,\subbase) \ra \subbase$, such that the map

\[(\pi_\subbase^\algebr,d_\normal\pi_\algebr): \normal(\algebr,\algebr|_{\subbase}) \xra{\sim} \algebr|_{\subbase} \ltimes \normal(\base,\subbase)\]

is an isomorphism of Lie algebroids. Moreover, $\mathbb{P}_{\pi}(\algebr,\algebr|_{\subbase})=\mathbb{P}(\algebr,\algebr|_{\subbase})$, and the above map descends to an isomorphism of Lie algebroids $\mathbb{P}(\algebr,\algebr|_{\subbase}) \xra{\sim} \algebr|_{\subbase} \ltimes \mathbb{P}(\base,\subbase)$.

§.§.§ Blow-up of a foliation

As a next example, we will show that foliations behave well under the blow-up construction with respect to a leaf. We fix a pair of manifolds $(M,L)$ such that $M$ is equipped with a foliation $\mathcal{F}$ for which $L \subset M$ is a leaf, i.e. $L$ is a leaf of the foliation $\mathcal{F}$ that is a closed embedded submanifold (reminder: closed in the topological sense).
[Blow-up of a foliated manifold]

We will show here that, for a closed embedded leaf $L \subset M$, $\blup(M,L)$ naturally inherits a foliation.

The Lie algebroid $\pi: T\mathcal{F} \ra M$, whose anchor map $\iota: T\mathcal{F} \hookrightarrow TM$ is the inclusion, has the Lie subalgebroid $T\mathcal{F}|_{L} = TL$ (note: $\Gamma(T\mathcal{F},TL) = T\mathcal{F}$). It suffices now to show that the Lie algebroid

\[\blup(T\mathcal{F},TL) \ra \blup(M,L)\]

(recall Proposition <ref>)
again has an injective anchor map. This anchor map is given by

\[\blup(T\mathcal{F},TL) \xra{\blup(\iota)} \blup_{\pi_{TM}}(TM,TL) \xra{\widetilde\anchor} T\blup(M,L).\]

Since $T\mathcal{F} \cap TL = TL$, we see that, by Proposition <ref>, the map

\[\blup(\iota): \blup_{\pi}(T\mathcal{F},TL) \ra \blup_{\pi_{TM}}(TM,TL),\]

is an injective immersion. However, the map

\[\widetilde\anchor: \blup_{\pi_{TM}}(TM,TL) \ra T\blup(M,L)\]

has, over a point $[y,\xi] \in \mathbb{P}(M,L)$, a $1$-dimensional kernel (see Example <ref>). Therefore, we have to show that, fiberwise, the image of $\blup(\iota)|_{\mathbb{P}(M,L)}$ has trivial intersection with the kernel of $\widetilde\anchor|_{\mathbb{P}(M,L)}$. To show this, we can work in a vector bundle chart, so we can consider $\mathcal{F}$ to be a trivial foliation on $M=\mathbb{R}^n$, with $L=\mathbb{R}^p \times \{0\}$. The quotient map $q: \normal(M,L) \setminus 0_N \cong \mathbb{R}^{p} \times (\mathbb{R}^{q} \setminus \{0\}) \rightarrowdbl \mathbb{P}(M,L) \cong \mathbb{R}^p \times \mathbb{RP}^{q-1}$ can locally be expressed as the map

\[\varphi_i \circ q: \mathbb{R}^p \times (\mathbb{R}^q\setminus \{0\}) \ra \mathbb{R}^p \times \mathbb{R}^{q-1} \textnormal{ given by } (y,\xi) \mapsto (y,\frac{\xi_1}{\xi_i},\dots,\frac{\xi_{i-1}}{\xi_i},\frac{\xi_{i+1}}{\xi_i},\dots,\frac{\xi_q}{\xi_i}),\]

where $(U_i^{\mathbb{RP}},\varphi_i)$ is a standard chart of $\mathbb{RP}^{q-1}$. Therefore, we can write (locally), for $(y,\xi) \in \mathbb{R}^{n}$,

\[dq(y,\xi)(\zeta,\eta)=(\zeta,\frac{\eta_1}{\xi_i} - \frac{\eta_i\xi_1}{\xi_i^2},\dots,\frac{\eta_{i-1}}{\xi_i} - \frac{\eta_i\xi_{i-1}}{\xi_i^2},\frac{\eta_{i+1}}{\xi_i} - \frac{\eta_{i+1}\xi_1}{\xi_i^2},\dots,\frac{\eta_q}{\xi_i} - \frac{\eta_i\xi_q}{\xi_i^2}).\]

Notice that this map has kernel given by

\[\{(0,\lambda \cdot \xi_1,\dots,%\lambda \cdot \xi_{i-1},\lambda \cdot \xi_i,\lambda \cdot \tfrac{\xi_{i+1}}{\xi_i},\dots,
\lambda \cdot \xi_q) \in \mathbb{R}^{p+q} \mid \lambda \in \mathbb{R}\}.\]

However, an element $(0,\lambda \cdot \xi_1,\dots,%\lambda \cdot \xi_{i-1},\lambda \cdot \xi_i,\lambda \cdot \tfrac{\xi_{i+1}}{\xi_i},\dots,
\lambda \cdot \xi_q) \in \mathbb{R}^{p+q}$ is tangent to $\mathbb{R}^p \times \{v\}$ for some $v \in \mathbb{R}^q$ if and only if $\lambda=0$. This shows that, indeed, fiberwise, the image of $\blup(\iota)|_{\mathbb{P}(M,L)}$ has trivial intersection with the kernel of $\widetilde\anchor|_{\mathbb{P}(M,L)}$ (note: $\widetilde\anchor([y,\xi]) = dq(y,\xi)$; see Proposition <ref>).

Alternatively, we can use Proposition <ref>: $L \subset M$ is saturated, so

\[\blup_{\pi}(T\mathcal{F},TL) = \blup(T\mathcal{F},TL),\]

and so that the statement can also be shown by using the commutativity of the following diagram:

(Tℱ,TL) [r,""] [d,"∼"]     T(M,L) [r,"d"]     TM [d,"="]

Tℱ ⋉(,)  [r,"_1"]     Tℱ [r,"ι",hook]     TM.

We denote the blow-up of the foliation $\mathcal{F}$ along $L$ by

\[\blup(\mathcal{F},L).\]

Similarly, the induced foliation on $\mathbb{P}(M,L)$ is written as $\mathbb{P}(\mathcal{F},L)$.

To describe the leaves of the blow-up foliation $\blup(\mathcal{F},L)$, recall Proposition <ref>. That is, in $M \setminus L \subset \blup(M,L)$ the partition into leaves is equal to the partition into leaves of $M \setminus L \subset M$, because $\blup(T\mathcal{F},TL)|_{M \setminus L} \cong T\mathcal{F}|_{M \setminus L}$.
In $\mathbb{P}(M,L) \subset \blup(M,L)$, the leaves are described by the subbundle:

\[\mathbb{P}(T\mathcal{F},TL) \ra \mathbb{P}(M,L)\]

of the vector bundle $\blup(T\mathcal{F},TL)$.
To be more specific about the nature of this foliation, we will use that $\mathbb{P}(M,L) \ra L$ is a fiber bundle with typical fiber $\mathbb{RP}^{q-1}$. The foliation $\mathbb{P}(\mathcal{F},L)$ falls into the following class of foliations:
Let $E \xra{\pi} \base$ be a fiber bundle with typical fiber $F$. A foliation $\mathcal{F}$ on $E$ is called transverse to the fibers if $\codim \mathcal{F} = \dim F$ (i.e. $\dim \mathcal{F} = \dim \base$), and for all leaves $L$ of $\mathcal{F}$, if $e \in L$, then the fiber $E_x$ of $E$ over $x\coloneqq\pi(e)$ is transverse to $L$, and $L \xra{\pi|_L} \base$ is a covering projection.

Since the fibers of $\mathbb{P}(M,L) \ra L$ are compact, the following result ensures that the foliation $\mathbb{P}(\mathcal{F},L)$ is indeed transverse to the fibers.
Let $E \xra{\pi} \base$ be a fiber bundle with typical fiber $F$, and let $\mathcal{F}$ be a foliation on $E$. If $F$ is compact, then $\mathcal{F}$ is transverse to the fibers if and only if for all leaves $L$ of $\mathcal{F}$, if $e \in L$, then the fiber $E_x$ of $E$ over $x\coloneqq\pi(e)$ is transverse to $L$ (i.e. the assumption in Definition <ref> that $L \xra{\pi|_L} \base$ is a covering projection is redundant in this case).

Obviously, we only have to prove the “if” statement. The idea is as follows: each $x \in \base$ admits an open neighbourhood $U$ such that, for all $u \in U$, $E_u \cap L'$ consists of precisely one point for all leaves $L'$ of the foliation $\mathcal{F}|_{\pi^{-1}(U)}$, whose leaves are given by the connected components of the intersection of leaves in $\mathcal{F}$ with $\pi^{-1}(U)$. Then, for every leaf $L'$ of $\mathcal{F}|_{\pi^{-1}(U)}$, we will see that $\pi_{L'}: L' \ra U$ is a bijection, and therefore a diffeomorphism, which shows that $\pi|_L: L \ra \base$ is a covering projection (note: since $\dim L = \dim \base$, $\pi|_L: L \ra \base$ is a local diffeomorphism).

Since the fibers are compact, we can reduce the problem to showing that, for all $x \in \base$,

\begin{align}\label{eq: compact fibers claim}
&\textnormal{if } e \in E_x, \textnormal{ there is a foliation chart } (W,\varphi) \textnormal{ around } e \textnormal{ such that, for all } w \in \pi(W), \nonumber \\
&\textnormal{the intersection of } E_w \textnormal{ with a plaque of } W \textnormal{ consists of exactly one point}.
\end{align}

To see why the statement follows from this claim, notice that, since $E_x$ is compact, we can cover $E_x$ by finitely many charts $(W_1,\varphi_1),\dots,(W_k,\varphi_k)$ with the mentioned property in (<ref>). So, by choosing a connected open set

\[U \subset \bigcap_{i=1}^k \pi(W_k)\]

which contains $x$, if $L'$ is a leaf of $\mathcal{F}|_{\pi^{-1}(U)}$, it is contained in a plaque $L_1'$ of, say, $W_1$ (note: $\pi(L_1') = \pi(W_1) \supset U$ by (<ref>)). In particular, $\pi|_{L'}: L' \ra U$ is injective, and therefore also surjective, since

\[\pi(L') = \pi(L_1' \cap \pi^{-1}(U)) = \pi(L_1') \cap \pi(\pi^{-1}(U)) = U.\]

This shows that $\pi|_{L'}: L' \ra U$ is a bijective local diffeomorphism, i.e. it is a diffeomorphism.

It remains to prove (<ref>). To do this, let $L$ be a leaf through $e \in E_x$.
We can find a connected open subset $V \ni e$ of $L$ such that, for all $v \in V$, $E_{\pi(v)} \cap V = \{v\}$. Then the map

\[\pi^{-1}(\pi(V)) \ra V \textnormal{ given by } z \mapsto v; \quad v \in E_{\pi(z)} \cap V\]

is well-defined, and it maps $V \subset \pi^{-1}(\pi(V))$ onto $V$. Moreover, by shrinking $V$, and arguing locally, the map is readily verified to be a submersion. Again, by shrinking $V$, we may assume that there is a foliation chart $(W',\varphi')$ such that $\overline{V}$ is contained in a plaque of $W'$, and so we may pick a foliation chart $(W,\varphi)$ such that $W \subset W' \cap \pi^{-1}(\pi(V))$ and $\varphi(V \cap W) = \varphi(W) \cap (\mathbb{R}^p \times \{0\})$ (where $p \coloneqq \dim L$). The chart $(W,\varphi)$ has the desired property from (<ref>). This proves the statement.

To see that $\mathbb{P}(\mathcal{F},L)$ is transverse to the fibers, we have to prove, by the above result, that the leaves are transverse to the fibers. Since this is a local question, we may reduce to the case that $\mathcal{F}$ is the trivial foliation on $M=\mathbb{R}^n$, with $L=\mathbb{R}^p \times \{0\}$. But then $\mathbb{P}(M,L) \cong \mathbb{P}(\mathbb{R}^q,0) \times \mathbb{R}^p \cong \mathbb{RP}^{q-1} \times \mathbb{R}^p$, in which case it is obvious that the leaves are transverse to the fibers.

The existence of a foliation on a fiber bundle, which is transverse to the fibers, is determined by whether or not the fiber bundle is isomorphic (as fiber bundles) to a fiber bundle with a discrete structure group. Explicitly, having a discrete structure group translates into the following definition:
Let $E \xra{\pi} \base$ be a fiber bundle with typical fiber $F$. If we can find a fiber bundle atlas $\{(U_i,\psi_i)\}$ such that the maps

\[U_{ij} \ra \textnormal{Diff } F \textnormal{ given by } x \mapsto \psi_{ij}(x,\cdot): F \xra{\sim} F\]

(here, $U_{ij} \coloneqq U_i \cap U_j$ and $\psi_{ij} \coloneqq \psi_i \circ \psi_j^{-1}: U_{ij} \times F \xra{\sim} U_{ij} \times F$) are locally constant, then $E$ is said to have a discrete structure group.

If $E \xra{\pi} \base$ is a fiber bundle with typical fiber $F$, we see that $E$ admits a foliation whose plaques are given by the fibers of the map

\[\pi^{-1}(U_i) \xra[\psi_i]{\sim} U_i \times F \xra{\pr_2} F\]

(where $\{(U_i,\psi_i)\}$ is an atlas) if and only if $E$ has a discrete structure group (such that the condition from Definition <ref> is satisfied with the atlas $\{(U_i,\psi_i)\}$). Indeed, if we pick charts $(U_i,\varphi_i)$ of $\base$, and charts $(W_r,\chi_r)$ of $F$, then the maps

\[f \coloneqq \pr_2 \circ (\varphi_\ell \times \chi_s) \circ \psi_{ij} \circ (\varphi_k \times \chi_r)^{-1}: \varphi_k(U_{ijk\ell}) \times \chi_r(W_{rs}) \ra \chi_s(W_{rs})\]

satisfy $f(x,y)=f(y)$ if and only if the maps $x \mapsto \psi_{ij}(x,\cdot)$ are locally constant. It is readily verified that this foliation is transverse to the fibers.

We can also describe the foliation on $\mathbb{P}(M,L)$ by factoring through a foliation on $\normal(M,L)$. That the foliation $\mathcal{F}$ on $M$ induces a foliation on $\normal(M,L)$, we again use that the inclusion $\iota: T\mathcal{F} \hookrightarrow TM$ induces the injective anchor map

\[\normal(T\mathcal{F},TL) \xhookrightarrow{d_\normal\iota} \normal(TM,TL) \cong T\normal(M,L)\]

of the Lie algebroid $\normal(T\mathcal{F},TL) \ra \normal(M,L)$. This foliation is denoted by $\normal(\mathcal{F},L)$.
We can apply the above idea to show that $\normal(\mathcal{F},L)$ is a foliation transverse to the leaves (and similarly for $\mathbb{P}(M,L) \ra L$). Indeed, if $\{(U_i,\varphi_i)\}=\{(U_i,y_i^1,\dots,y_i^p,x_i^1,\dots,x_i^q)\}$ is an adapted chart (to $L$), recall that then we have the vector bundle coordinates

\[\normal(U_i,V_i) \ra V_i \times \mathbb{R}^q \textnormal{ given by } (y,\xi) \mapsto (y,dx_i^1(y)\xi,\dots,dx_i^q(y)\xi)\]

(with $V_i \coloneqq U_i \cap L$).

So, if $\{(U_i,\varphi_i)\}$ is a foliation atlas (with $0 \in \varphi_i(U_i)$),

then showing that $\normal(M,L) \ra L$ has a transverse foliation structure comes down to the fact that the maps

\[y \mapsto (d(x_i^1 \circ \varphi_j^{-1})(y)\cdot,\dots,d(x_i^q \circ \varphi_j^{-1})(y)\cdot) = (d(x_i^1 \circ \varphi_j^{-1})(0)\cdot,\dots,d(x_i^q \circ \varphi_j^{-1})(0)\cdot)\]

are locally constant (see Remark <ref>).
This shows that $\normal(M,L) \ra L$ naturally inherits a foliation which is transverse to the fibers. We denote this foliation by $\normal(\mathcal{F},L)$. If $\iota: T\mathcal{F} \hookrightarrow TM$ is the inclusion, then the corresponding integrable subbundle $T\normal(\mathcal{F},L)$ of $T\normal(M,L)$ is $\normal(T\mathcal{F},TL)$.
We can similarly prove that $\mathbb{P}(M,L) \ra L$ inherits a transverse foliation structure this way, and the corresponding integrable subbundle $T\mathbb{P}(\mathcal{F},L)$ of $T\mathbb{P}(M,L)$ is $\mathbb{P}(T\mathcal{F},TL)$.

The $\mathbb{R}^\times$-action on $\normal(M,L) \setminus 0_L$ leaves the foliation $\normal(\mathcal{F},L)|_{\normal(M,L) \setminus 0_L}$ invariant; that is, the diffeomorphisms

\[\lambda: \normal(M,L) \setminus 0_L \xra{\sim} \normal(M,L) \setminus 0_L \textnormal{ given by } (y,\xi) \mapsto (y,\lambda\xi)\]

map leaves to leaves (we can reduce to the local case, where this property of the foliation is obvious).
One can show now that the foliation descends to a foliation on the quotient $(\normal(M,L) \setminus 0_L)/\mathbb{R}^\times = \mathbb{P}(M,L)$, and it is precisely the foliation $\mathbb{P}(\mathcal{F},L)$.

Notice that the Lie groupoids

\begin{align*}
\blup(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L) &\rra \blup(M,L) \textnormal{ and } \\
\blup(\Hol(M,\mathcal{F}),\Hol(M,\mathcal{F})|_L) &\rra \blup(M,L)
\end{align*}

integrate the Lie algebroid $\blup(T\mathcal{F},TL) \ra \blup(M,L)$. We will show that the former groupoid is canonically isomorphic to the monodromy groupoid of the foliation $\blup(\mathcal{F},L)$.
We will use the following general fact about Lie groupoid-integrations of foliations.

Let $\group \rra M$ be an $\source$-connected groupoid, and assume it is an integration of the Lie algebroid $T\mathcal{F} \hookrightarrow TM$ of the foliated manifold $(M,\mathcal{F})$. Then there is a sequence of morphism of Lie groupoids, which are surjective local diffeomorphisms and integrate the identity map $\id_{T\mathcal{F}}$,

\[\Mon(M,\mathcal{F}) \xrightarrowdbl{F_\group^\Mon} \group \xrightarrowdbl{F_\group^\Hol} \Hol(M,\mathcal{F}),\]

such that the composition $F_\group^\Hol \circ F_\group^\Mon$ is the canonical map $F$ from Remark <ref>. Moreover, the map $F_\group^\Mon$ is an isomorphism of groupoids if and only if $\group$ is, in addition, $\source$-simply connected.

Since $\Mon(M,\mathcal{F})$ is $\source$-connected and $\source$-simply connected (since a source fiber is the universal covering space of a leaf; see Proposition <ref>), the map $F_\group^\Mon$ is obtained by using Proposition <ref>, and the same proposition provides an inverse precisely when $\group$ is $\source$-simply connected. It remains to prove that there is a morphism of Lie groupoids $F_\group^\Hol$ fitting into the diagram

(M,ℱ) [rd,"F", twoheadrightarrow][rr,"F_^",twoheadrightarrow]         [ld,"F_^",twoheadrightarrow]

But, by the surjectivity of these maps, there is only one candidate for the map $\group \rightarrowdbl \Hol(M,\mathcal{F})$, namely $F_\group^\Mon(g) \mapsto F(g)$.
This map is well-defined and surjective, but we will not prove it here; see e.g. <cit.>.

We have a canonical Lie groupoid isomorphism

\begin{align*}
\Mon(\blup(M,L),\blup(\mathcal{F},L)) &\xra{\sim} \blup(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L)
%\textnormal{ and } \\
%\Hol(\blup(M,L),\blup(\mathcal{F},L)) &\xra{\sim} \blup(\Hol(M,\mathcal{F}),\Hol(M,\mathcal{F})|_L)
\end{align*}

which restricts to an isomorphism

\begin{align*}
\Mon(\mathbb{P}(M,L),\mathbb{P}(\mathcal{F},L)) &\xra{\sim} \mathbb{P}(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L)
%\textnormal{ and } \\
%\Hol(\mathbb{P}(M,L),\mathbb{P}(\mathcal{F},L)) &\xra{\sim} \mathbb{P}(\Hol(M,\mathcal{F}),\Hol(M,\mathcal{F})|_L)
\end{align*}

(recall Proposition <ref>).

Since the groupoid

\[\blup(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L) \rra \blup(M,L)\]

is also an integration of the Lie algebroid $T\blup(\mathcal{F},L) \ra \blup(M,L)$, we only have to prove, by Lemma <ref>, that $\blup(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L)$ is $\source$-connected and $\source$-simply connected. This follows from Proposition <ref>, because it says that

\begin{equation}\label{eq: blup Mon saturated iso}
\blup(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L) \cong \Mon(M,\mathcal{F}) \ltimes \blup(\base,\subbase)
\end{equation}

which is $\source$-connected and $\source$-simply connected because $\Mon(M,\mathcal{F})$ is.

This proves the statement.

Notice that the diagram

((M,L),(ℱ,L)) [d,"F_(ℱ,L)",twoheadrightarrow] [r,"∼"]     ((M,ℱ),(M,ℱ)|_L) [d,"(F_ℱ)",twoheadrightarrow]

((M,L),(ℱ,L)) [r]     ((M,ℱ),(M,ℱ)|_L)

Similarly, we have the following statement.

We can describe a foliation $\DNC(\mathcal{F},L)$ on $\DNC(M,L)$, which induces, for all $t \in \mathbb{R}^\times$, the foliation $\mathcal{F}$ on $\DNC(M,L)_t \cong M$, and the foliation $\normal(\mathcal{F},L)$ on $\DNC(M,L)_0 \cong \normal(M,L)$. Moreover, we have a canonical Lie groupoid isomorphism

\begin{align*}
\Mon(\DNC(M,L),\DNC(\mathcal{F},L)) &\xra{\sim} \DNC(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L)
%\textnormal{ and } \\
%\Hol(\DNC(M,L),\DNC(\mathcal{F},L)) &\xra{\sim} \DNC(\Hol(M,\mathcal{F}),\Hol(M,\mathcal{F})|_L)
\end{align*}

that restricts to a Lie groupoid isomorphism

\begin{align*}
\Mon(\normal(M,L),\normal(\mathcal{F},L)) &\xra{\sim} \normal(\Mon(M,\mathcal{F}),\Mon(M,\mathcal{F})|_L).
%\textnormal{ and } \\
%\Hol(\normal(M,L),\normal(\mathcal{F},L)) &\xra{\sim} \normal(\Hol(M,\mathcal{F}),\Hol(M,\mathcal{F})|_L).
\end{align*}

If $\iota: T\mathcal{F} \hookrightarrow TM$ is the inclusion, then the foliation $\DNC(\mathcal{F},L)$ is described by the Lie algebroid

\[\DNC(T\mathcal{F},TL) \xhookrightarrow{\DNC(\iota)} \DNC(TM,TL) \hookrightarrow T\DNC(M,L).\]

That the foliation restricts to the foliation $\mathcal{F}$ on $\DNC(M,L)_t \cong M$, and the foliation $\normal(\mathcal{F},L)$ on $\DNC(M,L)_0 \cong \normal(M,L)$, follows simply from the fact that $\DNC(T\mathcal{F},TL)$ has $\DNC(T\mathcal{F},TL)_t \cong T\mathcal{F}$ and $\DNC(T\mathcal{F},TL)_0 \cong \normal(T\mathcal{F},TL)$ as subalgebroids over the bases $\DNC(M,L)_t \cong M$ and $\DNC(M,L)_0 \cong \normal(M,L)$, respectively.

The last statement is proven similarly as in Proposition <ref>

§.§.§ Other examples of blow-ups

Here we will describe a few more examples of blow-ups of Lie groupoids and Lie algebroids.
[Principal bundles]
Suppose $P \xra{\pi} \basetwo$ is a principal $G$-bundle, and let $Q \subset P$ be invariant with respect to the action of $G$; notice that then $Q \ra \pi(Q) \eqqcolon \subbasetwo$ is again a principal $G$-bundle. The action of $G$ on $\blup(P,Q)$ (see Corollary <ref>) is again free and proper (e.g. because the blow-down map $\blup(P,Q) \ra P$ is proper). This yields a principal bundle

\[\blup(P,Q) \ra \blup(P,Q)/G.\]

Let us also discuss the Gauge groupoid $\blup(P,Q) \otimes_G \blup(P,Q)$ and the Atiyah algebroid $A(\blup(P,Q))$.
We have an isomorphism

\[\blup_{\source,\target}((P \times P)/G, (Q \times Q)/G) \xra{\sim} \blup_{\source,\target}(P \times P, Q \times Q)/G,\]

and by Example <ref>, we obtain an isomorphism

\[\blup_{\source,\target}((P \times P)/G, (Q \times Q)/G) \xra{\sim} (\hat P \times_{\mathbb{R}} \hat P/\mathbb{R}^\times)/G,\]

where $\hat P = \DNC(P,Q) \setminus Q \times \mathbb{R}$. Similarly, we have an isomorphism

\[\blup_{\pi}(TP/G,TQ/G) \xra{\sim} (\ker d\hat{t}|_{\hat{P}}/\mathbb{R}^\times)/G\]

(see Example <ref>).
The map $(\hat{P} \times_{\mathbb{R}} \hat{P})/\mathbb{R}^\times \ra \blup(P,Q) \times \blup(P,Q)$ now induces a map

\[\blup_{\source,\target}((P \times P)/G,(Q \times Q)/G) \ra (\blup(P,Q) \times \blup(P,Q))/G,\]

and the map $\ker d\hat{t}|_{\hat{P}}/\mathbb{R}^\times \ra T\blup(P.Q)$ induces a map $\blup_{\pi}(TP/G,TQ/G) \ra T\blup(P,Q)/G$. Lastly, notice that $\blup_{\source,\target}((P \times P)/G, (Q \times Q)/G)$ and $\blup_{\pi}(TP/G,TQ/G)$ are not transitive.

A remark in between: if $\groupoid$ is a Lie groupoid such that $\identity(\base) \subset \group$ is closed (i.e. $\group$ is Hausdorff; see Proposition <ref>) with Lie algebroid $\algebroid$, then

\[\mathbb{P}(\group,\identity(\base)) = (\normal(\group,\base) \setminus 0_\base)/\mathbb{R}^\times \cong (\algebr \setminus 0_\base)/\mathbb{R}^\times = \mathbb{P}(\algebr),\]

so we can think of $\blup(\group,\identity(\base))$ as “replacing $\identity(\base) \subset \group$ with the projectivisation of the Lie algebroid of $\group$” (that is, as a manifold; note: $\blup_{\source,\target}(\group,\identity(\base))= \emptyset$).

We will examine now what we can say about the blow-up of pullback Lie groupoids/algebroids.

Let $(\groupoid,\subgroupoid)$ be a pair of Lie groupoids and let $f: (\basetwo,\subbasetwo) \ra (\base,\subbase)$ be a smooth map of pairs such that $f^{-1}(\subbase)=\subbasetwo$ and $d_\normal f$ is fiberwise injective. Moreover, assume that $f \times f$ (seen as a map of pairs) has a clean intersection with the map $(\target,\source)$ (seen as a map of pairs $(\group,\subgroup) \ra (\base \times \base, \subbase \times \subbase)$), and that $\blup(f) \times \blup(f)$ has a clean intersection with $(\blup(\target),\blup(\source))$. Then

\[\blup_{\source^!,\target^!}(f^!\group,f^!\subgroup) \xra{\sim} \blup(f)^!\blup_{\source,\target}(\group,\subgroup)\]

as Lie groupoids (we denoted $\source^!$ and $\target^!$ for the source and target map of $f^!\group$, respectively).

By assumption, the pull-back groupoids $f^!\group$, $f^!\subgroup$ and $\blup(f)^!\blup_{\source,\target}(\group,\subgroup)$ are Lie groupoids. Notice that, by viewing $f^!\group$ and $f^!\subgroup$ as subgroupoids of $\basetwo \times \group \times \basetwo$, we can view $(f^!\group,f^!\subgroup)$ as a pair of Lie groupoids (notice that $f^!\subgroup \subset f^!\group$ is indeed closed). From Proposition <ref>, we see that

\begin{align*}
\blup_{\source^!,\target^!}(f^!\group,f^!\subgroup) &\xra{\sim} \blup(\basetwo,\subbasetwo) \tensor[_{\blup(f)}]{\times}{_{\blup(\source)}} \blup_{\source,\target}(\group,\subgroup) \tensor[_{\blup(\target)}]{\times}{_{\blup(f)}} \blup(\basetwo,\subbasetwo) \\
&= \blup(f)^!\blup_{\source,\target}(\group,\subgroup),
\end{align*}

(observe that the source of this map is correct $\blup_{\source^!,\target^!}(f^!\group,f^!\subgroup)=\blup_{f \times (\target,\source) \times f}(f^!\group,f^!\subgroup)$),
and it is readily verified to be a morphism of Lie groupoids. This proves the statement.

Following a similar reasoning, we can prove a statement for pullbacks of Lie algebroids as well.

Let $(\algebroid,\subalgebroid)$ be a pair of Lie algebroids and let $f: (\basetwo,\subbasetwo) \ra (\base,\subbase)$ be a smooth map of pairs such that $f^{-1}(\subbase)=\subbasetwo$ and $d_\normal f$ is fiberwise injective. Moreover, assume that $\anchor_{\algebr}$ (seen as a map of pairs) has a clean intersection with the map $df$ (seen as a map of pairs), and that $\anchor_{\blup_{\pi}(\algebr,\subalgebr)}$ has a clean intersection with $d\blup(f)$. Then

\[\blup_{\pi^!}(f^!\algebr,f^!\subalgebr) \xra{\sim} \blup(f)^!\blup_{\pi}(\algebr,\subalgebr)\]

as Lie algebroids (we denoted $\pi^!$ for the projection map of $\pi^!\algebr$).

By assumption, the pull-back algebroids $f^!\algebr$, $f^!\subalgebr$ and $\blup(f)^!\blup_{\pi}(\algebr,\subalgebr)$ are Lie algebroids. We can view $(f^!\algebr,f^!\subalgebr)$ as a pair of Lie algebroids (notice that $f^!\subalgebr \subset f^!\algebr$ is indeed closed). From Proposition <ref>, we see that

\begin{align*}
\blup_{\pi^!}(f^!\algebr,f^!\subalgebr) &\xra{\sim} \blup_{\pi_{T\base}}(T\basetwo,T\subbasetwo) \tensor[_{\blup(df)}]{\times}{_{\blup(\anchor)}} \blup_{\pi}(\algebr,\subalgebr) \\
&\xra{\sim} T\blup(\basetwo,\subbasetwo) \tensor[_{d\blup(f)}]{\times}{_{\anchor_{\blup_{\pi}(\algebr,\subalgebr)}}} \blup_{\pi}(\algebr,\subalgebr) \\
&= \blup(f)^!\blup_{\pi}(\algebr,\subalgebr),
\end{align*}

where we used in the second isomorphism that the diagram

_π_T(T,T) [r,"(df)"] [d,""]     _π_T(T,T) [d,""]     _π(,) [l,"()"] [d, "="]

T(,) [r,"d(f)"]     T(,)     _π(,) [l,"__π(,)"]

The above map is readily verified to be an isomorphism of Lie algebroids. This proves the statement.

§.§ Blow-up of differentiable stacks

In the proofs presented in this section, it is useful to use a different characterisation of Morita equivalences. We need the following definition.<cit.>
Let $\groupoid$ and $\grouptwoid$ be Lie groupoids. A morphism of Lie groupoids $\group \ra \grouptwo$ is called a Morita morphism if the base map $f$ is a surjective submersion, and the map

\[\varphi \times (\target_{\group},\source_{\group}): \group \ra \grouptwo \tensor[_{(\target_{\grouptwo},\source_{\grouptwo})}]{\times}{_{f \times f}} (\base \times \base)\]

is a diffeomorphism. If $(\groupoid,\subgroupoid)$ and $(\grouptwoid,\subgrouptwoid)$ are pairs of Lie groupoids, then a Lie groupoid morphism $\varphi: (\group,\subgroup) \ra (\grouptwo,\subgrouptwo)$ of pairs is called a Morita morphism of pairs if $\varphi: \group \ra \grouptwo$ and $\varphi|_{\subgroup}: \subgroup \ra \subgrouptwo$ are Morita morphisms. Additionally, the base map $f$ of $\varphi$ is required to satisfy $f^{-1}(\subbasetwo)=\subbase$.

The reason we put the extra condition on the base map $f$ is to ensure that, if we apply the blow-up functor, then $\blup(f)$ is a map $\blup(\base,\subbase) \ra \blup(\basetwo,\subbasetwo)$. This condition suffices, because $f$ is a surjective submersion, so $f: \base \ra \basetwo$ has a clean intersection with $\subbasetwo$, and therefore $d_\normal f$ is also fiberwise injective (see Remark <ref>).

Two groupoids $\groupoid$ and $\grouptwoid$ are Morita equivalent if and only if there is a groupoid $\mathcal{P} \rra U$ and Morita morphisms $\mathcal{P} \ra \group$ and $\mathcal{P} \ra \grouptwo$.

This suggest the following definition for pairs of Lie groupoids:

A Morita equivalence between pairs of Lie groupoids $(\groupoid,\subgroupoid)$ and $(\grouptwoid,\subgrouptwoid)$ is given by a pair of Lie groupoids $(\mathcal{P} \rra U, \mathcal{Q} \rra V)$ together with Morita morphisms $(\mathcal{P},\mathcal{Q}) \ra (\group,\subgroup)$ and $(\mathcal{P},\mathcal{Q}) \ra (\grouptwo,\subgrouptwo)$. In this case, $(\groupoid,\subgroupoid)$ and $(\grouptwoid,\subgrouptwoid)$ are said to be Morita equivalent; we write $(\groupoid,\subgroupoid) \sim (\grouptwoid,\subgrouptwoid)$.

Using Proposition <ref>, we can describe another way to define Morita equivalence between pairs of groupoids. Namely, two pairs of groupoids $(\groupoid,\subgroupoid)$ and $(\grouptwoid,\subgrouptwoid)$ are Morita equivalent if and only if there is a pair of manifolds $(E,F)$ together with a Morita equivalence

\[\base \xla{\pi_\base} E \xra{\pi_\basetwo} \basetwo\]

such that $F=\pi_\base^{-1}(\subbase)=\pi_\basetwo^{-1}(\subbasetwo)$, and

\[\subbase \xla{\pi_\base|_{F}} F \xra{\pi_\basetwo|_{F}} \subbasetwo\]

is also a Morita equivalence.

As in the case of the usual notion of Morita equivalence, Morita equivalence in the setting of pairs of groupoids is an equivalence relation.

By appropriately defining pairs of differentiable stacks, we can construct the deformation to the normal cone and blow-up of differentiable stacks.

A pair of differentiable stacks is a pair $(\mathfrak{X},\mathfrak{Y})$, where $\mathfrak{X}=[\groupoid]$ and $\mathfrak{Y}=[\grouptwoid]$ are differentiable stacks, together with a morphism $\mathfrak{Y} \ra \mathfrak{X}$ given by

\[(\grouptwoid) \sim (\subgroupoid) \xra{\iota} (\groupoid),\]

where $\iota$ is an embedding of Lie groupoids.

We will now show that, as claimed, the constructions of deformation to the normal cone, and of blow-up, are “stable” under Morita equivalence.

Let $(\groupoid,\subgroupoid)$ and $(\grouptwoid,\subgrouptwoid))$ be Lie groupoids which are Morita equivalent. Then $\DNC(\groupoid,\subgroupoid)$ and $\DNC(\grouptwoid,\subgrouptwoid)$ are Morita equivalent.

It suffices to show that, if $F: (\group,\subgroup) \ra (\grouptwo,\subgrouptwo)$ is a Morita morphism of pairs, then $\DNC(F)$ is a Morita morphism. And indeed, the base map of $\DNC(F)$ is $\DNC(f): \DNC(\base,\subbase) \ra \DNC(\basetwo,\subbasetwo)$, which is a surjective submersion by Proposition <ref>. That the map

\[\DNC(\varphi) \times (\DNC(\target_{\group}),\DNC(\source_{\group})): \DNC(\group,\subgroup) \ra \DNC(\grouptwo,\subgrouptwo) \tensor[_{(\DNC(\target_{\grouptwo}),\DNC(\source_{\grouptwo}))}]{\times}{_{\DNC(f) \times \DNC(f)}} \DNC(\base,\subbase) \times \DNC(\base,\subbase)\]

is a diffeomorphism goes as follows: notice that $(\DNC(\target_{\grouptwo})(g),\DNC(\source_{\grouptwo})(g))=(\DNC(f)(z),\DNC(f)(w))$ can only happen if $(z,w) \in \DNC(\base,\subbase) \times_{\mathbb{R}} \DNC(\base,\subbase)$ (because $(\DNC(\target_{\grouptwo})(g),\DNC(\source_{\grouptwo})(g))$ is an element of $\DNC(\basetwo,\subbasetwo) \times_{\mathbb{R}} \DNC(\basetwo,\subbasetwo)$). Therefore, by Proposition <ref>, we can equivalently prove that

\[\DNC(\varphi) \times \DNC(\target_{\group},\source_{\group}): \DNC(\group,\subgroup) \ra \DNC(\grouptwo,\subgrouptwo) \tensor[_{\DNC(\target_{\grouptwo},\source_{\grouptwo})}]{\times}{_{\DNC(f \times f)}} \DNC(\base \times \base, \subbase \times \subbase)\]

is a diffeomorphism. The result follows by again using Proposition <ref>, and the observation that the map $\DNC(\varphi \times (\target_{\group},\source_{\group}))$ is a diffeomorphism.

Let $(\groupoid,\subgroupoid)$ and $(\grouptwoid,\subgrouptwoid))$ be Lie groupoids which are Morita equivalent. Then $\blup_{\source_\group,\target_\group}(\group,\subgroup)$ and $\blup_{\source_\grouptwo,\target_\grouptwo}(\grouptwo,\subgrouptwo)$ are Morita equivalent.

Again, it suffices to prove that, if $F: (\group,\subgroup) \ra (\grouptwo,\subgrouptwo)$ is a Morita morphism of pairs, then $\blup(F)$ is a Morita morphism. Notice that, by Proposition <ref> and Remark <ref>, $\blup(F)$ is a morphism of Lie groupoids. The map $\blup(f): \blup(\base,\subbase) \ra \blup(\basetwo,\subbasetwo)$ is a surjective submersion by Proposition <ref>. To see in this case that the map $\blup(\varphi) \times (\blup(\target_{\group}),\blup(\source_{\group}))$ is a diffeomorphism, we use
Proposition <ref> and the fact that $\blup(\varphi \times (\target_{\group},\source_{\group}))$ is a diffeomorphism. This proves the statement.

From this we see that the blow-up of a pair of differentiable stacks is well-defined and yields another differentiable stack.

A special type of differentiable stack is a so-called orbifold. To describe, in this way, what an orbifold is, we need to know what a proper groupoid is.
A groupoid $\groupoid$ is called proper if the map

\[(\target,\source): \group \ra \base \times \base\]

is a proper map.

Observe that a Lie group $G$ (a groupoid over a point) is proper if and only if $G$ is compact, and an action groupoid $G \ltimes M$ is proper if and only if the action is proper. One can show that properness is a Morita invariant property.

An orbifold $\mathfrak{O}$ is a differentiable stack for which there exists a representative groupoid $\groupoid$ which is étale (i.e. $\dim \group = \dim \base$) and proper. We will call a groupoid $\groupoid$ an orbifold groupoid if it is étale and proper.

There are more concrete descriptions of orbifolds (see e.g. <cit.>), but we will not go into this. The property of being étale is not Morita invariant. One can show that a groupoid is Morita equivalent to an étale one if and only if the groupoid is a so-called folaited groupoid, which means that all the isotropy groups are discrete (this property is Morita invariant, and from this we also see that being étale is not a Morita invariant property; see <cit.>).

We now have the following result.

Let $\groupoid$ be a Lie groupoid and let $\subbase \subset \base$ be saturated. If $\groupoid$ is (a) étale, or (b) proper, then

\[\blup(\group,\group|_\subbase) \rra \blup(\base,\subbase)\]

is again (a) étale or (b) proper. In particular we can blow up orbifold groupoids, along this special class of sub-orbifold groupoids represented by Lie subgroupoids of the form $\group|_\subbase \rra \subbase$ (with $\subbase \subset \base$ saturated), and obtain another orbifold groupoid.

(a) Since $\subbase \neq \base$, we have $\dim \blup(\group,\group|_\subbase) = \dim \group$ and $\dim \blup(\base,\subbase) = \dim \base$.

(b) Consider the following commutative diagram:

(,|_) rr((),()) d         (,) ×(,) d×

rr(,)         ×.

Since $(\target,\source)$ is proper by assumption, and the blow-down map of a manifold is proper (see Proposition <ref>), we see, by the above commutative diagram, that the map $(\bldown \times \bldown) \circ (\blup(\target),\blup(\source))$ is proper. The result now follows from Lemma <ref> (if $g \circ f$ is proper, then $f$ is proper).

Since $\group|_\subbase \subset \group$, also in the non-étale case, has the same codimension as $\subbase \subset \base$, we see that $\group|_\subbase \rra \subbase$ is again étale. Moreover, $(\target,\source): \group|_\subbase \ra \subbase \times \subbase$ is proper, because it equals the map

\[\group|_\subbase \xhookrightarrow{\iota} \group \xra{(\target,\source)} \base \times \base\]

which is the composition of proper maps (note: $\iota$ is a proper embedding precisely because $\group|_\subbase \subset \group$ is a closed embedded submanifold). This proves the statement.

This yields a blow-up construction (at least in special cases) for orbifolds using the blow-up construction for differentiable stacks described in this section. For more details on suborbifolds, and how to describe them using the theory of Lie groupoids, see <cit.>.