Condensed Mathematics
The internal Hom of condensed sets and condensed abelian groups and a
prismatic construction of the real numbers

Rodrigo Marlasca Aparicio

Part B Extended Essay
Merton College, University of Oxford
MMath Mathematics
Trinity Term 2021

###### Contents

  1. 1 Introduction
  2. 2 Condensed objects
    1. 2.1 Preliminaries
    2. 2.2 Condensed abelian groups
    3. 2.3 Resolutions of condensed abelian groups
    4. 2.4 Condensed cohomology
  3. 3 Enriched Hom functors
    1. 3.1 Internal Hom for condensed sets
    2. 3.2 Internal Hom for condensed abelian groups
    3. 3.3 The derived category of condensed abelian groups
  4. 4 Derived mapping spaces
    1. 4.1 The spectral sequence argument
    2. 4.2 A few important cases
    3. 4.3 Locally compact abelian groups
    4. 4.4 RHom in locally compact abelian groups
    5. 4.5 Examples
  5. 5 A prismatic construction of the real numbers
    1. 5.1 The main result
    2. 5.2 A helpful ring
    3. 5.3 Proof of the main result

##  1 Introduction

Condensed mathematics are a very recent area of study, first proposed by
Dustin Clausen and Peter Scholze. They provide a generalization of topological
spaces, groups, rings and modules that is arguably better behaved than the
usual categories.

One of the most common motivating examples is the map
$\mathbb{R}^{\text{disc}}\to\mathbb{R}$ in the category of topological abelian
groups. This is clearly an epimorphism and a monomorphism, so if we had an
abelian category we would expect an isomorphism of topological groups.
However, this is not the case due to the different topologies, making many
constructions difficult in this category. If we consider this as a map of
condensed abelian groups, it can be shown that it is not an epimorphism, so
the cokernel is not trivial.

Having an abelian category of topological groups has many advantages. Mainly,
one can start using many homological methods like extension and torsion
groups. For example, while continuous group cohomology does not always give
long exact sequences, we can substitute these by extension groups in the
abelian category of condensed $\mathbb{Z}[G]$-modules, fixing this problem.

This new theory has yielded some surprising results and not only with
topological groups. An example is the Whitehead problem: is every abelian
group with $\operatorname{Ext}^{1}(A,\mathbb{Z})=0$ a free abelian group? This
question turns out to be independent of ZFC [She74], but it has been proved
that if
$\underline{\operatorname{Ext}}^{1}(\underline{A},\underline{\mathbb{Z}})=0$
then $A$ is free abelian, where the extension group is enriched over condensed
abelian groups [CS20].

Very recently, condensed mathematics have also been used to provide a more
categorical and abstract approach to functional analysis that has been coined
analytic geometry [Sch].

Other theories have been developed to solve the aforementioned issues. For
example, pyknotic objects [BH19] use the same core ideas but set theoretic
difficulties are solved differently. Finally, a different approach is given by
bornological modules [BBK21], which form a quasi-abelian category but avoid
some unintuitive problems of condensed sets like non-trivial spaces with the
empty set as its underlying set. However, this essay will solely focus on
condensed objects and some of their applications.

  * •

Section 2 provides an introduction to the study of condensed objects, paying
particular attention to condensed abelian groups and their resolutions and
cohomology.

  * •

In Section 3 we talk about closed symmetric monoidal structures in various
condensed categories. We define the internal Hom for condensed sets and
provide an alternative proof of [Sch19], Proposition 4.2.

  * •

Section 4 shows a few important examples of the enriched Hom in the derived
category.

  * •

Section 5 studies an intermediate step of a theorem of Clausen and Scholze,
constructing the real numbers from discrete spaces.

##  2 Condensed objects

###  2.1 Preliminaries

Sites give a more generalised version of a topological space. Instead of open
sets, we will have sieves.

######  Definition 2.1.

Let $\mathcal{C}$ be a category and let $c$ be an object of $\mathcal{C}$. A
_sieve_ $S:\mathcal{C}^{op}\to\text{Set}$ is a subfunctor of
$\text{Hom}(-,c)$.

Given a morphism $f:c^{\prime}\to c$ we can define the _pullback_ of $S$ by
$f$ as

$\displaystyle f^{*}S(c^{\prime\prime})=\\{g:c^{\prime\prime}\to
c^{\prime}|fg\in S(c^{\prime\prime})\\}$

That is, arrows that composed with $f$ give a map in the sieve.

######  Definition 2.2.

A _Grothendieck topology_ on a category $\mathcal{C}$ is a collection, for
each object $c$ of $\mathcal{C}$, of sieves called the _covering sieves_ of
$c$. They must satisfy the following axioms.

  1. 1.

If $S$ is a covering sieve on $X$ and $f:Y\to X$ is a morphism, then the
pullback $f^{*}S$ is a convering sieve on $Y$.

  2. 2.

If $S$ is a covering sieve on $X$, $T$ is a sieve on $X$ and for any $f\in S$
the sieve $f^{*}T$ is a covering sieve, then $T$ is a covering sieve.

  3. 3.

$\operatorname{Hom}(-,X)$ is a covering sieve for any $X$ in $\mathcal{C}$.

A _site_ is a category $\mathcal{C}$ together with a Grothendieck topology.

######  Remark 2.3.

We make the distinction between a topology and a pretopology. A _covering
family_ is a collection of maps with a common codomain.

A _Grothendieck pretopology_ is a collection of covering families such that

  1. 1.

For all objects $X$ of $\mathcal{C}$, all maps $Y\to X$ and all covering
families $\\{X_{i}\to X\\}$, all the relevant fiber products exist and the
collecion $\\{X_{i}\times_{X}Y\to Y\\}$ is a covering family.

  2. 2.

If $\\{X_{i}\to X\\}$ is a covering family and for every $i$ the collection
$\\{X_{ij}\to X_{i}\\}$ is also a covering family, the composition
$\\{X_{ij}\to X_{i}\to X\\}$ is a covering family.

  3. 3.

If $f$ is an isomorphism, $\\{f\\}$ is a covering family.

Every pretopology can be extended to a topology by considering all sieves
containing a covering family [AGV72, Remark 1.3.1].

We may also define a sheaf on a site. Remember that if we consider the
category of posets of a topological space $X$, if $U,V\subset X$ are open then
$U\cap V=U\times_{X}V$, which gives rise to the following natural definition.

######  Definition 2.4.

An $X$-valued _sheaf_ on a site $\mathcal{C}$ is a functor
$F:\mathcal{C}^{op}\to X$ such that for any $X\in\mathcal{C}$ and any cover
$\\{X_{i}\to X\\}$ the following diagram is an equaliser

$\displaystyle
F(X)\xhookrightarrow{}\prod_{i}F(X_{i})\rightrightarrows\prod_{i,j}F(X_{i}\times_{X}X_{j})$

where the stacked arrows correspond to the induced maps from the fiber product
to $X_{i}$ and $X_{j}$.

######  Remark 2.5.

The inclusion $F(X)\xhookrightarrow{}\prod_{i}F(X_{i})$ is the local character
of the sheaf, since the value of $F(X)$ is determined by the value at a cover.
On the other hand, the equaliser corresponds to the gluing axiom, because
sections coinciding at the fiber product assemble together to form the section
at $X$.

We now define a particularly important site for the study of condensed
objects.

######  Definition 2.6.

A _profinite set_ is a Hausdorff, compact and totally disconnected space.

The _pro-étale site_ $*_{\text{proét}}$ of a point is the category of
profinite sets $S$ whose covering families are finite sets of jointly
surjective maps.

######  Remark 2.7.

The above terminology comes from the pro-étale site of a scheme $X$, since the
site of profinite sets corresponds to the pro-étale site of $X$ when it is
just a single point [Del77].

To check this is a pretopology, observe that a limit of profinite spaces is
profinite [Sta21, Tag 0ET8], so all relevant fiber products exist.

For maps of topological spaces

${A}$${C}$${B}$$\scriptstyle{f}$$\scriptstyle{g}$

the fiber product $A\times_{C}B$ is given by $\\{(a,b)|f(a)=g(b)\\}\subseteq
A\times B$, and we have obvious projection maps to $A$ and $B$ [CM12].

Then, for a covering family $\\{X_{i}\to X\\}$ (so they are finite and jointly
surjective) and a map $f:Y\to X$, for every $y\in Y$, there is some $x$ in
some $X_{i}$ such that their images onto $X$ coincide. The fiber products then
surject onto $Y$ and pullback gives a new covering family.

The composition of two covering families is clearly surjective and any
isomorphism is surjective, so it forms a covering family. This shows all
pretopology axioms are satisfied.

One would want to naively define condensed sets as a sheaf on
$*_{\text{proét}}$. Evaluation at a profinite set can be thought of as the
maps from this profinite set into our condensed set. In this way, condensed
sets capture not only the topology of the space but also how it interacts with
other spaces.

However, the category of profinite sets is large and this would present many
set theoretic issues, like the category not being locally small. We use
extremally disconnected sets to find an equivalent definition.

######  Definition 2.8.

An _extremally disconnected set_ is a projective object in the category of
compact Hausdorff spaces.

######  Remark 2.9.

[Gle58] shows that the category of compact Hausdorff spaces has enough
projectives, so any compact Hausdorff $X$ admits a surjection
$S\twoheadrightarrow{}X$ where $S$ is extremally disconnected.

This implies that a sheaf on profinite sets is determined by the values at
extremally disconnected sets, and it can be shown that the category of sheaves
on profinite sets is equivalent to the category of functors on extremally
disconnected sets taking finite disjoint unions to products [Sch19,
Proposition 2.7].

Extremally disconnected sets are difficult to come by (the most common ones
are the Stone-Čech compactification of a discrete space) but their
characterisation as projectives will be enough for this essay.

######  Definition 2.10.

A condensed set/group/ring is a functor

$\displaystyle T:\\{\text{extremally disconnected
sets}\\}^{\text{op}}\to\\{\text{sets/groups/rings}\\}$

such that $T(\emptyset)=*$ and for any extremally disconnected $S_{1},S_{2}$,

$\displaystyle T(S_{1}\sqcup S_{2})=T(S_{1})\times T(S_{2})$

and is the left Kan extension of its restriction to $\kappa$-small extremally
disconnected sets, for some uncountable strong limit cardinal $\kappa$.

${\text{ED}^{\text{op}}}$${\text{ED}_{\kappa}^{\text{op}}}$${\\{\text{sets/groups/rings}\\}}$$\scriptstyle{T}$$\scriptstyle{T_{\kappa}}$

######  Remark 2.11.

Equivalently, they are also sheaves on profinite sets satisfying a similar Kan
extension condition. It will then make sense to talk about sheafification.

Informally, one can think of the left Kan condition as the sheaf being
determined by its values on $\kappa$-small extremally disconnected sets for
some $\kappa$. This is not exactly a category of sheaves on a site, but it
shares many of its features.

[Sch19, Remark 2.13] shows that this category is stable under the section-wise
formation of limits and filtered colimits, so these will be easily computed.

######  Proposition 2.12 ([Sch19], Proposition 2.15).

There is a functor from T1 topological spaces/groups/rings to condensed
sets/groups/rings sending $T\to\underline{T}$, where $T(S)=C(S,T)$ for
extremally disconnected $S$.

If $A$ is compactly generated as a topological space and $A,B$ are T1
topological spaces/groups/rings then

$\displaystyle\operatorname{Hom}(\underline{A},\underline{B})=\operatorname{Hom}(A,B)$

where the right hom-set is in the category of topological spaces/groups/rings.

######  Remark 2.13.

The T1 condition on the topology is to ensure that the defined sheaf is the
left Kan extension of the restriction to $\kappa$-small extremally
disconnected sets for some $\kappa$ ([Sch19] Proposition 2.15).

This functor allows us to think about condensed sets in a more natural way.
Let $T$ be a condensed set and let $S$ be extremally disconnected. Then,

$\displaystyle\operatorname{Hom}(\underline{S},T)=T(S)$

by the Yoneda lemma. Hence, evaluation at $S$ can be thought of as the set of
continuous maps from $S$ to $T$.

###  2.2 Condensed abelian groups

Condensed abelian groups turn out to be as well-behaved as one could
reasonably expect them to be.

######  Proposition 2.14 ([Sch19], Theorem 2.2).

The category of condensed abelian groups is abelian, satisfying the same
Grothendieck axioms as the category of groups. That is,

  1. 1.

(AB3) All colimits exist.

  2. 2.

(AB3*) All limits exist.

  3. 3.

(AB4) Arbitrary direct sums are exact.

  4. 4.

(AB4*) Arbitrary products are exact.

  5. 5.

(AB5) Arbitrary filtered colimits are exact.

  6. 6.

(AB6) For any index set $J$ and filtered categories $I_{j},j\in J$ with
functors $i\to M_{i}$ from $I_{j}$ to condensed abelian groups,

$\displaystyle\lim_{i_{j},\forall j\in J}\prod_{j\in J}M_{ij}\cong\prod_{j\in
J}\lim_{i_{j}\in I_{j}}M_{ij}$

######  Remark 2.15.

Since all limits and colimits of abelian groups commute with finite products
(=direct sums) we have that the category of functors on extremally
disconnected spaces taking finite disjoint unions to products is stable under
section-wise limits and colimits.

When passing to condensed sets, note that a left Kan extension can be written
as a filtered colimit, so it will commute with any colimits. As for limits,
[Sch19, Remark 2.13] shows that they also preseve the left Kan condition by
choosing a suitable cardinal $\kappa$.

Limits and colimits being computed section-wise leads to naturally talking
about injections or surjections of condensed abelian groups (instead of
monomorphisms and epimorphisms). Unlike the usual category of sheaves, both
kernels and cokernels can be computed at every section. A map is a
monomorphism (resp. epimorphism) if and only if it is injective (resp.
surjective) at every section.

We highlight the contrast with condensed sets, where one can only take limits
and filtered colimits section-wise. The problem with a general colimit is that
they don’t commute with all products in the category of sets so the sheaf
condition might not be preserved.

There is a condensed analogue to free abelian groups.

######  Proposition 2.16 ([Sch19], Theorem 2.2).

There is a left adjoint functor to the forgetful functor from condensed
abelian groups to condensed sets, $T\mapsto\mathbb{Z}[T]$. This is the
sheafification of $S\mapsto\mathbb{Z}[T(S)]$.

For extremally disconnected $S$, the free condensed abelian group
$\mathbb{Z}[\underline{S}]$ is projective.

######  Remark 2.17.

Using this adjunction, we see that for extremally disconnected $S$ and a
condensed abelian group $M$,

$\displaystyle\operatorname{Hom}(\mathbb{Z}[\underline{S}],M)\cong\operatorname{Hom}(\underline{S},M)\cong
M(S)$

where the last isomorphism is by the Yoneda lemma.

The functor from T1 topological abelian groups is not exact or even left
exact, but we can focus on certain special cases.

######  Proposition 2.18.

Suppose

${0}$${A}$${B}$${C}$${0}$$\scriptstyle{f}$$\scriptstyle{g}$

is a sequence of Hausdorff topological abelian groups that is exact as a
sequence of abelian groups, $f$ is a closed map and for every compact
$K\subseteq C$ there is a compact $B^{\prime}\subseteq C$ such that
$g(B^{\prime})\supseteq K$. Then, the induced condensed sequence

${0}$${\underline{A}}$${\underline{B}}$${\underline{C}}$${0}$

is exact.

###### Proof.

Applying the functor and analysing at every section, it is enough to show that

$\displaystyle
0\xrightarrow{}C(S,A)\xrightarrow{f\circ}C(S,B)\xrightarrow{g\circ}C(S,C)\xrightarrow{}0$

is exact for every extremally disconnected set $S$. Injectivity of the first
map is clear.

Suppose $h\in C(S,B)$ and $gh=0$. Then, $h(S)\subseteq\ker g=f(A)$. Define
$l:S\to A$ by $l=f^{-1}\circ h$. This is continuous because $f$ is a closed
map and $f\circ l=h$. This shows
$\operatorname{Im}(f\circ)\supseteq\ker(g\circ)$.

Since $g\circ f=0$, the reverse inclusion is clear.

Finally, we must show that $C(S,B)$ surjects onto $C(S,C)$. Consider a map
$h:S\to C$. Then, $h(S)$ is compact so for some compact $B^{\prime}\subseteq
B$, $g(B^{\prime})\supseteq h(S)$. The image $h(S)$ is a compact subset of a
Hausdorff space so it is closed and $g^{-1}(h(S))$ is closed.

$B$ is Hausdorff as well, so $B^{\prime}$ is closed and $g^{-1}(h(S))\cap
B^{\prime}$ (a closed subset of a compact space) must be compact as well.
$g^{-1}(h(S))\cap B^{\prime}$ surjects onto $h(S)$ and $S$ is projective in
the category of compact Hausdorff spaces, so $h$ lifts to a map $S\to
g^{-1}(h(S))\cap B^{\prime}\subseteq B$. ∎

######  Proposition 2.19.

The functor $M\mapsto\underline{M}$ commutes with direct sums of discrete
groups, where we endow the direct sum of groups with the discrete topology.

###### Proof.

Let $\\{M_{i}\\}$ be a collection of discrete spaces and let $M=\bigoplus
M_{i}$. Looking at the problem section-wise, it suffices to prove that

$\displaystyle\bigoplus C(S,M_{i})\cong C(S,M)$

where $M$ has the discrete topology. Let $f\in C(S,M)$. The space $S$ is
compact so its image onto a discrete space must be finite. It is clear then
that $f\in\bigoplus C(S,M_{i})$ by taking the projection into each coordinate.

Let $g\in\bigoplus C(S,M_{i})$. We want to prove that the induced map $S\to M$
is continuous. Let $g_{i}$ be the $i$th coordinate of $g$. Then, $g_{i}\neq 0$
for finitely many $i$. Each $g_{i}:S\to M_{i}$ partitions $S$ into clopen sets
(by taking preimages of singletons) and there must be finitely many of them by
compactness. By taking intersections, we can refine this to a finite partition
of clopen sets such that $g_{i}$ is constant on each of the clopen sets for
all $i$. This induces a continuous map $S\to\bigoplus M_{i}$. ∎

######  Proposition 2.20.

Let $(K_{i})_{i\in I}$ be a collection of compact Hausdorff spaces. Then,

$\displaystyle\underline{\prod_{I}K_{i}}=\prod_{I}\underline{K_{i}}$

######  Remark 2.21.

This applies whether the $K_{i}$ are topological spaces or groups.

###### Proof.

For extremally disconnected $S$, it is enough to show that

$\displaystyle C(S,\prod_{I}K_{i})=\prod_{I}C(S,K_{i})$

Let $p_{j}:\prod_{I}K_{i}\to K_{j}$ be the projection map. Let
$f:S\to\prod_{I}K_{i}$. Then, $(p_{i}\circ f)_{i\in I}\in\prod_{I}C(S,K_{i})$.

Now let $(g_{i})_{i\in I}\in\prod_{I}C(S,K_{i})$. We claim that the natural
function $g:S\to\prod_{I}K_{i}$ is continuous.

Let $U\subseteq\prod_{I}K_{i}$ be closed. Then,

$\displaystyle g^{-1}(U)=\bigcap_{i\in I}(g_{i}^{-1}\circ p_{i})(U)$

$p_{i}$ is a surjection from a compact space to a Hausdorff space, so it is a
closed map. Hence, $p_{i}(U)$ is closed. By continuity of $g_{i}$, the
preimage $(g_{i}^{-1}\circ p_{i})(U)$ is closed and hence the intersection is
closed. This concludes the proof. ∎

###  2.3 Resolutions of condensed abelian groups

Derived functors and the derived category will be built through projectives,
since the category of condensed abelian groups has enough projectives ([Sch19]
Theorem 2.2) but not any non-zero injectives (there is no published proof in
the literature of this fact, but Peter Scholze provided a proof in a Math
Overflow thread).

Projectives $\mathbb{Z}[\underline{S}]$ for an extremally disconnected set $S$
behave in a similar way to the projectives in the category of abelian groups
and we will use this to build certain resolutions.

######  Theorem 2.22 ([Sch19], Theorem 4.5).

Let $A$ be an abelian group. Then, there is a projective resolution of $A$

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}\mathbb{Z}[A^{r_{i,j}}]\to\cdots\to\bigoplus_{j=1}^{n_{1}}\mathbb{Z}[A^{r_{1,j}}]\to
A\to 0$

that is functorial in $A$ and the $r_{i,j}$ are independent of the choice of
$A$.

######  Remark 2.23.

We do not provide a description of the differentials but this will not be
necessary for any of our computations. The first few can be found in [BBM82]
and are of the form

$\displaystyle\mathbb{Z}[A^{4}]\oplus\mathbb{Z}[A^{3}]\oplus\mathbb{Z}[A^{3}]\oplus\mathbb{Z}[A^{2}]\oplus\mathbb{Z}[A]\to\mathbb{Z}[A^{3}]\oplus\mathbb{Z}[A^{2}]\to\mathbb{Z}[A^{2}]\to\mathbb{Z}[A]\to
A$

for an arbitrary abelian group $A$ and [Sch19, Theorem 4.10] shows that this
may be extended to a resolution of the desired form.

######  Theorem 2.24.

Let $M$ be a condensed abelian group and let $r_{i,j}$ be defined as above.
Then, there is a resolution of $M$ of the form

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}\mathbb{Z}[M^{r_{i,j}}]\to\cdots\to\bigoplus_{j=1}^{n_{1}}\mathbb{Z}[M^{r_{1,j}}]\to
M\to 0$

######  Remark 2.25.

This resolution is not necessarily projective, since the objects
$\mathbb{Z}[M^{r_{i,j}}]$ might not be projective. $\mathbb{Z}[\underline{S}]$
is projective for an extremally disconnected set $S$ but $\mathbb{Z}[T]$ might
not be for a general condensed set $T$.

###### Proof.

Define $Z[M]$ as the presheaf $S\mapsto\mathbb{Z}[M(S)]$, so that the
sheafification of $Z[M]$ is $\mathbb{Z}[M]$. Then, for any extremally
disconnected set we have a resolution

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}Z[(M(S))^{r_{i,j}}]\to\cdots\to\bigoplus_{j=1}^{n_{1}}Z[(M(S))^{r_{1,j}}]\to
M(S)\to 0$

$M\mapsto M(S)$ commutes with limits and colimits so we get a resolution

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}Z[M^{r_{i,j}}](S)\to\cdots\to\bigoplus_{j=1}^{n_{1}}Z[M^{r_{1,j}}](S)\to
M(S)\to 0$

The functoriality of the resolution ensures that for every map $T\to S$ (and
hence a map $M(S)\to M(T)$) of extremally disconnected sets, the resolutions
of $M(S)$ and $M(T)$ form a commutative ladder diagram, so each of the maps
induces a natural transformation. We get a sequence of presheaves

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}Z[M^{r_{i,j}}]\to\cdots\to\bigoplus_{j=1}^{n_{1}}Z[M^{r_{1,j}}]\to
M\to 0$

This sequence is exact at every section so the sequence must be exact. We may
now sheafify and we are done because sheafification is an exact functor. ∎

It is possible to explicitly construct the projective resolution of a
condensed abelian group of the form $\mathbb{Z}[\underline{S}]$, where $S$ is
a compact Hausdorff space.

######  Theorem 2.26.

Let $S$ be a compact Hausdorff space and let $S_{\bullet}\to S$ be a
simplicial hypercover of $S$ by extremally disconnected spaces. Then,

$\displaystyle\dots\to\mathbb{Z}[\underline{S_{1}}]\to\mathbb{Z}[\underline{S_{0}}]\to\mathbb{Z}[\underline{S}]\to
0$

is a projective resolution of $\mathbb{Z}[\underline{S}]$.

######  Remark 2.27.

By condensing and applying the free condensed abelian group functor, we would
get a simplicial condensed abelian group rather than a complex. However, we
obtain the desired complex by extracting the Moore complex in the same way as
the Dold-Kan correspondence functor.

###### Proof.

This is a standard fact about the category of abelian sheaves on a site
[Sta21, Tag 01GF].

Hence, the complex is exact. Finally, the $\mathbb{Z}[\underline{S_{i}}]$ are
projective by 2.16. ∎

###  2.4 Condensed cohomology

We will be working with different kinds of cohomology throughout this essay,
which will in some cases coincide.

######  Definition 2.28.

Let $S$ be a compact Hausdorff space. Then, we have the following ways of
computing the cohomology $H^{i}(S,M)$ for some discrete group $M$.

  1. 1.

Sheaf cohomology: the right derived functor of
$\Gamma:\operatorname{Ab}(S)\to\operatorname{Ab}$ defined by $\Gamma(F)=F(S)$.
Then, $H^{i}_{\text{sheaf}}(S,M)=R^{i}\Gamma(M)$, where $M$ is considered as
the constant sheaf with value $M$.

  2. 2.

Čech cohomology: referred to as $H^{i}_{\text{\v{C}ech}}(S,M)$.

They are isomorphic for compact Hausdorff $S$ [God, Theorem 5.10].

Condensed abelian groups allow us to give a third definition of cohomology.

######  Definition 2.29.

For a T1 space $S$ and an abelian group $M$, we define condensed cohomology as

$\displaystyle
H^{i}(S,M)=\operatorname{Ext}^{i}_{\operatorname{Cond(Ab)}}(\mathbb{Z}[\underline{S}],\underline{M})$

where $M$ is given the discrete topology.

We can also write more generally for a condensed abelian group $A$ and a
condensed set $T$,

$\displaystyle H^{i}(T,A)=\operatorname{Ext}^{i}(\mathbb{Z}[T],A)$

######  Remark 2.30.

By Theorem 2.26, we have a projective resolution of the condensed abelian
group $\mathbb{Z}[\underline{S}]$, for a compact Hausdorff $S$, using a
simplicial hypercover $S_{\bullet}\to S$. The condensed cohomology
$H^{i}(S,M)$ would then be computed by the complex
$\operatorname{Hom}(\mathbb{Z}[\underline{S_{i}}],M)\cong M(S_{i})$ by 2.16.
This means that the complex

$\displaystyle 0\to M(S_{0})\to M(S_{1})\to M(S_{2})\to\dots$

has the groups $H^{i}(S,M)$ as its cohomology.

######  Theorem 2.31 ([Sch19], Theorem 3.2, 3.3).

Let $S$ be a compact Hausdorff space. Then,

  1. 1.

$H^{i}(S,M)\cong H^{i}_{\text{\v{C}ech}}(S,M)$

  2. 2.

$H^{i}(S,\mathbb{R})\cong 0$ for $i>0$ and $H^{0}(S,\mathbb{R})\cong
C(S,\mathbb{R})$.

##  3 Enriched Hom functors

###  3.1 Internal Hom for condensed sets

The category of sheaves on a site $\mathcal{C}$ naturally has an enriched hom
functor by considering

$\displaystyle{\mathscr{H}\kern-0.5ptom}(\mathcal{F},\mathcal{G})(S)=\operatorname{Hom}(\mathcal{F}|_{\mathcal{C}_{/S}},\mathcal{G}|_{\mathcal{C}_{/S}})$

where $\mathcal{C}_{/S}$ is the slice category. We provide the following more
practical definition. A discussion of the equivalence between definitions in
the presheaf category can be found in [na21].

In this subsection we study this construction for condensed sets as well as
some applications.

######  Proposition 3.1.

The category of condensed sets has a closed symmetric monoidal structure, with
the usual product as tensor product and internal hom given by

$\displaystyle\mathscr{H}\kern-0.5ptom(A,B)(S)=\operatorname{Hom}_{\operatorname{Cond(Set)}}(A\times\underline{S},B)$

###### Proof.

This could be proved in terms of an abstract adjoint functor theorem, but
instead we provide an explicit construction of the isomorphism that will be
necessary in a later result.

Let $M,N,P$ be condensed sets and let $g:P\to\mathscr{H}\kern-0.5ptom(M,N)$.
For extremally disconnected $S$, $(p,m)\in P(S)\times M(S)$ and
$\operatorname{id}:S\to S$ the identity function, we can define

$\displaystyle f_{S}(p,m)=(g_{S}(p))_{S}(m\times\operatorname{id})$

where the subscript of an extremally disconnected set on a sheaf morphism
indicates the induced map on sections. This is a map $P\times M\to N$.

In a more schematic way, this can be understood as

$\displaystyle g\mapsto(p\times m\mapsto(g(p)(m,\operatorname{id})))$

Let $f:P\times M\to N$. For extremally disconnected $S,T$, $p\in P(S)$ and
$(m,s)\in M(T)\times C(T,S)$.

$\displaystyle(g_{S}(p))_{T}(m,s)=f(s^{*}(p),m)$

where $s^{*}$ is the induced map $P(S)\to P(T)$. This is a map
$P\to\mathscr{H}\kern-0.5ptom(M,N)$.

More schematically,

$\displaystyle f\mapsto(p\mapsto(m\times s\mapsto f(s^{*}(p),m)))$

We claim that the above induces an isomorphism of hom sets.

On one side we have for $f:P\times M\to N$, by composing the two above
constructions,

$\displaystyle f\mapsto(p\times m\mapsto f(\operatorname{id}^{*}(p),m))$

But clearly $\operatorname{id}^{*}(p)=p$ so this gives

$\displaystyle f\mapsto(p\times m\mapsto f(p,m))=f$

In the other direction, we have

$\displaystyle g\mapsto(p\mapsto(m\times s\mapsto
g(s^{*}(p))(m,\operatorname{id})))$

We only need to check that $g(s^{*}(p))(m,\operatorname{id})=g(p)(m,s)$.

Writing this more precisely by making the extremally disconnected sets
explicit, we need to prove
$(g_{T}(s^{*}(p)))_{T}(m,\operatorname{id})=(g_{S}(p))_{T}(m,s)$ for a
continuous map $s:T\to S$. We will abuse notation and write $s^{*}$ for the
two maps $P(S)\to P(T)$ and
$\operatorname{Hom}(M\times\underline{S},N)\to\operatorname{Hom}(M\times\underline{T},N)$.
By the definition of a natural transformation we have a commutative diagram

${P(S)}$${\operatorname{Hom}(M\times\underline{S},N)}$${P(T)}$${\operatorname{Hom}(M\times\underline{T},N)}$$\scriptstyle{g}$$\scriptstyle{s^{*}}$$\scriptstyle{s^{*}}$$\scriptstyle{g}$

This tells us that
$(g_{T}(s^{*}(p)))_{T}(m,\operatorname{id})=(g_{S}(p))_{T}s^{*}(m,\operatorname{id})$.
Looking at the problem at the given section, we have $s^{*}$ inducing a map
$M(T)\times C(T,T)\to M(T)\times C(T,S)$. This is clearly given by composition
with $s$ so it follows that

$\displaystyle(g_{T}(s^{*}(p)))_{T}(m,\operatorname{id})=(g_{S}(p))_{T}s^{*}(m,\operatorname{id})=(g_{S}(p))_{T}(m,s)$

as required. ∎

The following gives us a more direct way of computing this enriched hom
functor.

######  Proposition 3.2.

Let $A$ and $B$ be Hausdorff topological spaces, where $A$ is compactly
generated. Then,

$\displaystyle\mathscr{H}\kern-0.5ptom(\underline{A},\underline{B})=\underline{C(A,B)}$

where $C(A,B)$ is given the compact-open topology.

######  Remark 3.3.

We include a reminder that for spaces $X$ and $Y$, the compact-open topology
on $C(X,Y)$ is defined as having subbasis the following sets. For $K\subseteq
X$ compact and $U\subseteq Y$ open,

$\displaystyle V(K,U)=\\{f\in C(X,Y)|f(K)\subseteq U\\}$

If $Y$ is locally compact and Hausdorff we get a hom-tensor adjunction

$\displaystyle C(X\times Y,Z)\cong C(X,C(Y,Z))$

where $C(Y,Z)$ is given the compact-open topology [AGPS02].

###### Proof.

Taking $S$-valued points for extremally disconnected $S$, we need an
isomorphism

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Set)}}(\underline{A}\times\underline{S},\underline{B})\to
C(S,C(A,B))$

By the hom-tensor adjunction of the compact-open topology, $C(S,C(A,B))\cong
C(A\times S,B)$. This is then clear because condensation is fully faithful
when restricted to compactly generated spaces. ∎

###  3.2 Internal Hom for condensed abelian groups

We can make an analogous construction with condensed abelian groups. The
following is a general property of abelian sheaves on a site.

######  Proposition 3.4.

Condensed abelian groups $\operatorname{Cond(Ab)}$ form a closed symmetric
monoidal category, where the tensor product $A\otimes B$ is given by the
sheafification of

$\displaystyle S\mapsto A(S)\otimes B(S)$

We also have an enriched hom functor $\underline{\operatorname{Hom}}$
satisfying the adjunction

$\displaystyle\operatorname{Hom}(P,\underline{\operatorname{Hom}}(M,N))=\operatorname{Hom}(P\otimes
M,N)$

If we apply this to the free object $P=\mathbb{Z}[\underline{S}]$ for an
extremally disconnected set $S$,

$\displaystyle\underline{\operatorname{Hom}}(M,N)(S)=\operatorname{Hom}(\mathbb{Z}[\underline{S}]\otimes
M,N)$

which allows us to characterise $\underline{\operatorname{Hom}}$.

With these enriched functors, we can derive an enriched version of the
adjunction between the forgetful functor
$\operatorname{Cond(Ab)}\to\operatorname{Cond(Set)}$ and free condensed
abelian groups.

######  Proposition 3.5.

Let $A$ be a condensed set and $B$ a condensed abelian group. Then, there is
an isomorphism of condensed sets

$\displaystyle\underline{\operatorname{Hom}}(\mathbb{Z}[A],B)\cong\mathscr{H}\kern-0.5ptom(A,B)$

###### Proof.

Taking $S$-valued points, we need an isomorphism

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\mathbb{Z}[A\times\underline{S}],B)\to\operatorname{Hom}_{\operatorname{Cond(Set)}}(A\times\underline{S},B)$

But this is clear by the adjunction between free condensed abelian groups and
the forgetful functor $\operatorname{Cond(Ab)}\to\operatorname{Cond(Set)}$. ∎

The closed symmetric monoidal structure on condensed sets allows us to prove
the following result.

######  Theorem 3.6.

Let $A$ and $B$ be Hausdorff topological groups, with $A$ compactly generated.
Then,

$\displaystyle\underline{\operatorname{Hom}}(\underline{A},\underline{B})\cong\underline{\operatorname{Hom}_{\operatorname{Ab}}(A,B)}$

where $\operatorname{Hom}_{\operatorname{Ab}}(A,B)$ carries the compact-open
topology.

###### Proof.

Taking $S$-valued points, we want to construct isomorphisms

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A}\otimes\mathbb{Z}[\underline{S}],\underline{B})\to
C(S,\operatorname{Hom}_{\operatorname{Ab}}(A,B))$

Using the hom-tensor adjunction in condensed abelian groups,

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A}\otimes\mathbb{Z}[\underline{S}],\underline{B})\cong\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A},\underline{\operatorname{Hom}}(\mathbb{Z}[\underline{S}],\underline{B}))\xhookrightarrow{}\operatorname{Hom}_{\operatorname{Cond(Set)}}(\underline{A},\underline{\operatorname{Hom}}(\mathbb{Z}[\underline{S}],\underline{B}))$

Now we can apply 3.5 and have an inclusion

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A}\otimes\mathbb{Z}[\underline{S}],\underline{B})\xhookrightarrow{}\operatorname{Hom}_{\operatorname{Cond(Set)}}(\underline{A},\underline{\operatorname{Hom}}(\mathbb{Z}[\underline{S}],\underline{B}))\cong\operatorname{Hom}_{\operatorname{Cond(Set)}}(\underline{A},\mathscr{H}\kern-0.5ptom(\underline{S},\underline{B}))$

By the hom-tensor adjunction in condensed sets,

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Set)}}(\underline{A},\mathscr{H}\kern-0.5ptom(\underline{S},\underline{B}))\cong\operatorname{Hom}_{\operatorname{Cond(Set)}}(\underline{A\times
S},\underline{B})\cong C(A\times S,B)$

where the last isomorphism is due to the fully faithful embedding of compactly
generated spaces in condensed sets. Then, we have an inclusion

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A}\otimes\mathbb{Z}[\underline{S}],\underline{B})\xhookrightarrow{}C(A\times
S,B)$

We claim that the induced maps $A\times S\to B$ are additive in $A$. Let
$f:\underline{A}\otimes\mathbb{Z}[\underline{S}]\to\underline{B}$. By our
construction, this induces a map of condensed abelian groups
$\underline{A}\to\underline{\operatorname{Hom}}(\mathbb{Z}[\underline{S}],\underline{B})$,
which by the inclusion used is a map of condensed sets
$\hat{f}:\underline{A}\to\mathscr{H}\kern-0.5ptom(\underline{S},\underline{B})$
that is additive in $\underline{A}$ at each section.

By 3.1, this induces
$\hat{g}:\underline{A}\times\underline{S}\to\underline{B}$. We now check that
this is linear in $A$ at each section. The construction in 3.1 gives that for
profinite $T$, and $(a,s)\in\underline{A}(T)\times\underline{S}(T)$,

$\displaystyle\hat{g}_{T}(a,s)=(\hat{f}(a))_{T}(s\times id)$

Since $\hat{f}$ was linear in $a$, it is easy to see that $\hat{g}$ is too.

Hence, the induced maps are linear in $A$. By the hom-tensor adjunction of the
compact-open topology, we have continuous maps $S\to C(A,B)$ such that the
induced maps on $A$ are additive. Hence, the image must be contained in
$\operatorname{Hom}_{\operatorname{Ab}}(A,B)$. This shows an inclusion

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A}\otimes\mathbb{Z}[\underline{S}],\underline{B})\xhookrightarrow{}C(S,\operatorname{Hom}_{\operatorname{Ab}}(A,B))$

Now we must show surjectivity. Let
$g:S\to\operatorname{Hom}_{\operatorname{Ab}}(A,B)$. By the reverse
construction, this induces a map
$\hat{g}:\underline{A}\times\underline{S}\to\underline{B}$ that is additive in
$\underline{A}$ at every section.

Applying 3.1, this induces a map of condensed sets
$\hat{f}:\underline{A}\to\mathscr{H}\kern-0.5ptom(\underline{S},\underline{B})$,
such that for profinite $T_{1}$ and $T_{2}$, $a\in\underline{A}(T_{1})$ and
$(s,t)\in\underline{S}(T_{2})\times\underline{T_{1}}(T_{2})$,

$\displaystyle(\hat{f}_{T_{1}}(a))_{T_{2}}(s,t)=\hat{g}_{T_{2}}(t^{*}(a),s)$

where $t^{*}:\underline{A}(T_{1})\to\underline{A}(T_{2})$ is the induced map
by the sheaf. Since $\underline{A}$ is a condensed abelian group, $t^{*}$ must
be additive in $A$, so $\hat{f}$ must be additive by additivity of $\hat{g}$.

Hence, the map is linear on $\underline{A}$ at every section so this results
in a map of condensed abelian groups
$\underline{A}\to\underline{\operatorname{Hom}}(\mathbb{Z}[\underline{S}],\underline{B})$.
By adjunctions, this lifts to a map
$\underline{A}\otimes\mathbb{Z}[\underline{S}]\to\underline{B}$ as required. ∎

######  Remark 3.7.

The above is proved in [Sch19, Proposition 4.2] but we include an alternative
proof. In the original argument, adjuctions are used to prove that a map
$\underline{A}\otimes\mathbb{Z}[\underline{S}]\to\underline{B}$ induces a map
$A\times S\to B$. More precisely, the argument used in these notes is

$\displaystyle\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\underline{A}\otimes\mathbb{Z}[\underline{S}],\underline{B})\xhookrightarrow{}\operatorname{Hom}_{\operatorname{Cond(Ab)}}(\mathbb{Z}[\underline{A}\times\underline{S}],\underline{B})\cong
C(A\times S,B)$

However, it did not seem a priori clear to the author why this map should be
additive in $A$ to conclude a map
$S\to\operatorname{Hom}_{\operatorname{Ab}}(A,B)$. Due to the adjunctions
used, both in the tensor product and the free condensed abelian group, it is
not obvious that this additivity should be preserved. Sheafification makes the
explicit maps somewhat uncertain. This led to the construction of the internal
Hom in condensed sets for this essay.

###  3.3 The derived category of condensed abelian groups

With the previous constructions, we can also build a derived category. Within
the conventional theory of $1$-categories, it is not known whether
$D(\operatorname{Cond(Ab)})\cong\operatorname{Cond}(D(\operatorname{Ab}))$.
There is no reason why $\operatorname{Cond}(D(\operatorname{Ab}))$ should be
triangulated.

However, this can all be solved using $\infty$-categories. Sheaves of abelian
groups on the category of extremally disconnected sets are very well-behaved
since homology commutes with evaluation on extremally disconnected sets. This
results in a hypercomplete $(\infty,1)$-topos which gives an equivalence
between the desired categories [Lur18, Theorem 2.1.2.2].

Nevertheless, this advanced treatment will not be necessary for the purposes
of this essay and we will treat $D(\operatorname{Cond(Ab)})$ as the usual
triangulated category, which allows essentially the same computations.
Whenever we talk about distinguished triangles, the $\infty$-categorical
equivalent object is a cofiber sequence.

######  Theorem 3.8 ([Sch19]).

The derived category of condensed abelian groups $D(\operatorname{Cond(Ab)})$
is closed symmetric monoidal, where we have an adjunction between an internal
hom and a total tensor product

$\displaystyle\operatorname{Hom}(M\otimes^{L}N,P)\cong\operatorname{Hom}(M,R\underline{\operatorname{Hom}}(N,P))$

######  Example 3.9.

We will explain in more detail how to calculate
$R\underline{\operatorname{Hom}}$, as the derived functor of
$\underline{\operatorname{Hom}}^{\bullet}$ in the derived category. Let $A$
and $B$ be two bounded above cochain complexes of condensed abelian groups.

We now take the double complex
$\\{\underline{\operatorname{Hom}}(A^{p},B^{-q})\\}$ with maps $A^{p}\to
A^{p+1}$ and $B^{-q}\to B^{-q-1}$, which by the functoriality of
$\underline{\operatorname{Hom}}$ turn into differentials

$\displaystyle
d^{h}:\underline{\operatorname{Hom}}(A^{p},B^{-q})\to\underline{\operatorname{Hom}}(A^{p+1},B^{-q})$
$\displaystyle
d^{v}:\underline{\operatorname{Hom}}(A^{p},B^{-q})\to\underline{\operatorname{Hom}}(A^{p},B^{-q-1})$

(one needs to adjust the sign of $d^{v}$ multiplying by $(-1)^{p+q+1}$ to
ensure the antisymmetry of the differentials). Define

$\displaystyle\underline{\operatorname{Hom}}^{\bullet}(A,B)=\operatorname{Tot}^{\Pi}(\\{\underline{\operatorname{Hom}}(A^{p},B^{-q})\\})$

$\underline{\operatorname{Hom}}^{\bullet}(-,B)$ is a morphism of triangulated
categories (when considering the homotopy category) so we may construct the
total derived functor $R\underline{\operatorname{Hom}}(-,B)$. Due to the lack
of injectives in condensed abelian groups it is not possible to do an
analogous construction with $\underline{\operatorname{Hom}}^{\bullet}(A,-)$.

We then obtain a bifunctor
$R\underline{\operatorname{Hom}}:D(\operatorname{Cond(Ab)})^{op}\times
D(\operatorname{Cond(Ab)})\to D(\operatorname{Cond(Ab)})$. This is completely
analogous to the usual $R\operatorname{Hom}$ and details may be found in
[Wei97].

$\operatorname{Cond(Ab)}$ has enough projectives, so we can find a Cartan-
Eilenberg resolution of projectives of $A^{op}$ ($A$ considered as a chain
complex so it is bounded below) and take the chain complex
$\operatorname{Tot}^{\oplus}$, which is quasi-isomorphic to $A^{op}$ ([Wei97]
§5.7). Since $A^{op}$ is bounded below, its Cartan-Eilenberg resolution is a
first quadrant bicomplex, so when applying $\operatorname{Tot}^{\oplus}$ we
get a bounded below complex. We now take again the opposite cochain complex to
get a complex of projectives quasi-isomorphic to $A$, call it $P$. Then,

$\displaystyle
R\underline{\operatorname{Hom}}(A,B)=\underline{\operatorname{Hom}}^{\bullet}(P,B)$

However, this work will focus on computing
$R\underline{\operatorname{Hom}}(A,B)$ where $A$ and $B$ are condensed abelian
groups and we are just considering them as cochain complexes concentrated on
degree zero. Then, we may substitute $A$ by just a projective resolution.

Observe then that the Cartan-Eilenberg resolution would be concentrated on the
vertical axis, and $R\underline{\operatorname{Hom}}(A,B)$ would be the functor
$\underline{\operatorname{Hom}}(-,B)$ applied to the projective resolution of
$A$.

######  Remark 3.10.

The adjunction gives the usual properties of $R\operatorname{Hom}$

  1. 1.

$R\underline{\operatorname{Hom}}(\bigoplus A_{i},B)=\prod
R\underline{\operatorname{Hom}}(A_{i},B)$

  2. 2.

$R\underline{\operatorname{Hom}}(A,\prod B_{j})=\prod
R\underline{\operatorname{Hom}}(A,B_{j})$

##  4 Derived mapping spaces

In this section we explore the internal hom $R\underline{\operatorname{Hom}}$
of the derived category.

###  4.1 The spectral sequence argument

Finding a projective resolution of condensed abelian groups is not a trivial
calculation. That is why we provide this construction that uses spectral
sequences and condensed cohomology.

######  Theorem 4.1.

Let $A$ and $M$ be condensed abelian groups and $S$ extremally disconnected.
Then, there is a spectral sequence

$\displaystyle
E_{1}^{pq}=\prod_{j=1}^{n_{p}}H^{q}(A^{r_{p,j}}\times\underline{S},M)\implies\underline{\operatorname{Ext}}^{p+q}(A[0],M[0])(S)$

###### Proof.

By 3.4,

$\displaystyle\underline{\operatorname{Hom}}(A,M)(S)=\operatorname{Hom}(A\otimes\mathbb{Z}[\underline{S}],M)$

Then we must have

$\displaystyle
R\underline{\operatorname{Hom}}(A,M)(S)=R\operatorname{Hom}(A\otimes\mathbb{Z}[\underline{S}],M)$

so we will be interested in studying the latter object to understand the
sections of $R\underline{\operatorname{Hom}}(A,M)(S)$. This same reasoning
will apply to ext groups because evaluation at extremally disconnected sets
commutes with homology, so

$\displaystyle\underline{\operatorname{Ext}}(A,M)(S)=\underline{\operatorname{Ext}}(A\otimes\mathbb{Z}[\underline{S}],M)$

By Theorem 2.24, we may pick a resolution of the form

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}\mathbb{Z}[A^{r_{i,j}}]\to\cdots\to\bigoplus_{j=1}^{n_{1}}\mathbb{Z}[A^{r_{1,j}}]\to
A\to 0$

This is not necessarily projective, but we obtain a complex of free condensed
abelian groups quasi-isomorphic to $A[0]$.

Since $\mathbb{Z}[\underline{S}]$ is projective, it is flat so we may tensor
by it to obtain a resolution of the form

$\displaystyle\cdots\to\bigoplus_{j=1}^{n_{i}}\mathbb{Z}[A^{r_{i,j}}\times\underline{S}]\to\cdots\to\bigoplus_{j=1}^{n_{1}}\mathbb{Z}[A^{r_{1,j}}\times\underline{S}]\to
A\otimes\mathbb{Z}[\underline{S}]\to 0$

Consider a Cartan-Eilenberg resolution $P^{**}$ of the above complex so
$\operatorname{Tot}^{\oplus}(P^{**})$ is a projective resolution of
$A\otimes\mathbb{Z}[\underline{S}]$. By Example 3.9,

$\displaystyle
R\operatorname{Hom}(A\otimes\mathbb{Z}[\underline{S}],M)=\operatorname{Hom}(\operatorname{Tot}^{\oplus}(P^{**}),M)$

where we apply $\operatorname{Hom}(-,M)$ on each degree. However,
$\operatorname{Hom}(-,M)$ commutes with finite products so

$\displaystyle
R\operatorname{Hom}(A\otimes\mathbb{Z}[\underline{S}],M)=\operatorname{Tot}^{\oplus}\operatorname{Hom}(P^{**},M)$

To compute $\operatorname{Ext}^{i}(A\otimes\mathbb{Z}[\underline{S}],M)$ we
may just take the cohomology of the total complex. For this computation, we
will use spectral sequences.

Take the spectral sequence corresponding to the double complex, induced by the
trivial filtration on columns. We then have that the first iteration is equal
to the vertical cohomology. However, columns in a Cartan-Eilenberg resolution
are projective resolutions so

$\displaystyle
E_{1}^{pq}\cong\operatorname{Ext}^{q}(\bigoplus_{j=1}^{n_{p}}\mathbb{Z}[A^{r_{p,j}}\times\underline{S}],M)\cong\prod_{j=1}^{n_{p}}\operatorname{Ext}^{q}(\mathbb{Z}[A^{r_{p,j}}\times\underline{S}],M)\cong\prod_{j=1}^{n_{p}}H^{q}(A^{r_{p,j}}\times\underline{S},M)$

The spectral sequence converges to the cohomology of the total complex so

$\displaystyle
E_{1}^{pq}\implies\operatorname{Ext}^{q}(A\otimes\mathbb{Z}[\underline{S}],M)\cong\underline{\operatorname{Ext}}^{q}(A,M)(S)$

∎

###  4.2 A few important cases

The following result is analogous to the equivalent calculation in the usual
category of abelian groups.

######  Lemma 4.2.

For any condensed abelian group $M$,

$\displaystyle R\underline{\operatorname{Hom}}(\underline{\mathbb{Z}},M)=M[0]$

###### Proof.

First we will prove that

$\displaystyle\underline{\mathbb{Z}}=\mathbb{Z}[\underline{*}]$

Notice that $\underline{*}$ is the zero sheaf, so $\mathbb{Z}[\underline{*}]$
is the locally constant sheaf with value $\mathbb{Z}$, which is exactly the
sheaf of constant functions with codomain $\mathbb{Z}$, so we get the desired
equality.

Hence, $\underline{\mathbb{Z}}$ is projective. So to consider
$R\underline{\operatorname{Hom}}$ we only need to look at
$\underline{\operatorname{Hom}}(\underline{\mathbb{Z}},M)$. For an extremally
disconnected set $S$,

$\displaystyle\underline{\operatorname{Hom}}(\mathbb{Z}[\underline{*}],M)(S)=\operatorname{Hom}(\mathbb{Z}[\underline{*}]\otimes\mathbb{Z}[\underline{S}],M)=\operatorname{Hom}(\mathbb{Z}[\underline{*}\times\underline{S}],M)=\operatorname{Hom}(\mathbb{Z}[\underline{S}],M)=M(S)$

So equality follows. ∎

######  Theorem 4.3.

Let $M$ be a discrete abelian group. Then,
$R\underline{\operatorname{Hom}}(\underline{\mathbb{R}},\underline{M})=0$.

###### Proof.

By Theorem 4.1 we obtain spectral sequences for
$\underline{\operatorname{Ext}}^{q}(\underline{\mathbb{R}},\underline{M})$ and
$\underline{\operatorname{Ext}}^{q}(0,\underline{M})=0$. Hence, to prove the
result it would suffice to prove that the map

$\displaystyle H^{q}(\mathbb{R}^{r}\times S,M)\to H^{q}(S,M)$

induced by pullback of $0\times\operatorname{id}:S\to\mathbb{R}^{r}\times S$
is an isomorphism so they have isomorphic spectral sequences and hence
converge to the same cohomology.

We now may write $\mathbb{R}^{r}$ as the filtered colimit of compact spaces.

$\displaystyle\mathbb{R}^{r}=\lim_{\longrightarrow}[-n,n]^{r}$

We claim that this filtered colimit commutes with the condensation functor.

Looking at it section wise (since condensed sets are stable under pointwise
filtered colimits) it is enough to prove that
$C(T,\mathbb{R}^{r})=\operatorname{colim}C(T,[-n,n]^{r})$ for extremally
disconnected $T$. The space $T$ is compact, so any function
$T\to\mathbb{R}^{r}$ is contained in a compact set of the form $[-n,n]^{r}$
and equality follows.

The functor of free condensed abelian groups is a left adjoint, so it
preserves colimits. Hence,

$\displaystyle\mathbb{Z}[\underline{\mathbb{R}^{r}}]=\lim_{\longrightarrow}\mathbb{Z}[\underline{[-n,n]^{r}}]$

We now write this as a homotopy colimit. $H^{q}(\mathbb{R}^{r}\times S,M)$ is
the cohomology of
$R\operatorname{Hom}(\mathbb{Z}[\underline{\mathbb{R}^{r}}\times\underline{S}],\underline{S})$
and $R\operatorname{Hom}$ takes homotopy colimits to homotopy limits so

$\displaystyle
R\operatorname{Hom}(\mathbb{Z}[\underline{\mathbb{R}^{r}}\times\underline{S}],M)=\operatorname{holim}R\operatorname{Hom}(\mathbb{Z}[\underline{[-n,n]^{r}}\times\underline{S}],M)$

The cohomology of the complex on the right is $H^{q}([-n,n]^{r}\times,M)$ and
by pullback of $0\times\operatorname{id}:S\to[-n,n]^{r}\times S$ we get a map
on the cohomologies

$\displaystyle H^{q}([-n,n]^{r}\times S,M)\to H^{q}(S,M)$

This is an isomorphism for every $q$, since by Theorem 2.31 this corresponds
to sheaf cohomology which is homotopy invariant. Hence, the map is a quasi-
isomorphism between complexes. This is true for every $n$, so the homotopy
limit must also be a weak equivalence which gives us an isomorphism of
cohomologies. Then, the map

$\displaystyle H^{q}(\mathbb{R}^{r}\times S,M)\to H^{q}(S,M)$

is an isomorphism. ∎

######  Theorem 4.4 ([Sch19], Theorem 4.3).

Let $M$ be a discrete abelian group and let
$A=\prod_{I}\mathbb{R}/\mathbb{Z}$. Then,

$\displaystyle
R\underline{\operatorname{Hom}}(\underline{A},\underline{M})=\bigoplus_{I}\underline{M}[-1]$

###### Proof.

We have a short exact sequence

${0}$${\underline{\mathbb{Z}}}$${\underline{\mathbb{R}}}$${\underline{\mathbb{R}}/\underline{\mathbb{Z}}}$${0}$

because the corresponding short exact sequence of abelian groups satisfies the
hypothesis of 2.18.

Since the functor $R\underline{\operatorname{Hom}}(-,\underline{M})$ is a
morphism of triangulated categories,
$R\underline{\operatorname{Hom}}(\underline{\mathbb{Z}},\underline{M})=\underline{M}$
and $R\underline{\operatorname{Hom}}(\underline{\mathbb{R}},\underline{M})=0$,
we just complete the distinguished triangle and obtain

$\displaystyle
R\underline{\operatorname{Hom}}(\underline{\mathbb{R}/\mathbb{Z}},\underline{M})=\underline{M}[-1]$

It then would suffice to show that

$\displaystyle\lim_{\longrightarrow}R\underline{\operatorname{Hom}}(\prod_{J}\underline{\mathbb{R}/\mathbb{Z}},\underline{M})=R\underline{\operatorname{Hom}}(\prod_{I}\underline{\mathbb{R}/\mathbb{Z}},\underline{M})$

where $J$ runs over all finite subsets of $I$. Applying Theorem 4.1 to make an
argument with spectral sequences, it suffices to prove that the natural map

$\displaystyle\lim_{\longrightarrow}H^{i}(S\times\prod_{J}\mathbb{R}/\mathbb{Z},M)\to
H^{i}(S\times\prod_{I}\mathbb{R}/\mathbb{Z},M)$

is an isomorphism. But this follows from comparison with sheaf cohomology
[AGV72, Lemma 4.1.3]. ∎

######  Theorem 4.5 ([Sch19], Theorem 4.3).

Let $A$ be a compact abelian group. Then,

$\displaystyle
R\underline{\operatorname{Hom}}(\underline{A},\underline{\mathbb{R}})=0$

###### Proof.

Apply Theorem 4.1 to obtain a spectral sequence $E^{pq}$ with

$\displaystyle E^{p,q}_{1}=\prod_{j=1}^{n_{p}}H^{q}(A^{r_{p,j}}\times
S,\mathbb{R})\implies\underline{\operatorname{Ext}}^{p+q}(\underline{A},\underline{\mathbb{R}})(S)$

$A^{r_{p,j}}\times S$ is compact and by Theorem 2.31, $H^{q}(K,\mathbb{R})$
vanishes for compact $K$ and $q>0$, and $H^{0}(K,\mathbb{R})=C(K,\mathbb{R})$.
Hence, all of the spectral sequence vanishes except for the horizontal axis.
Then, it suffices to show that the spectral sequence vanishes entirely in the
next iteration. We will prove that the vertical axis

$\displaystyle 0\to\bigoplus_{j=1}^{n_{0}}C(A^{r_{0,j}}\times
S,\mathbb{R})\to\bigoplus_{j=1}^{n_{1}}C(A^{r_{1,j}}\times
S,\mathbb{R})\to\dots$

is exact.

Call $F(A)_{\bullet}$ the resolution of $A$ used to obtain the spectral
sequence. Let $2:F(A)\to F(A)$ be the chain map where every component is
multiplied by $2$, and let $[2]:F(A)\to F(A)$ be the map induced by
multiplying $A$ by $2$ (this is possible since the resolution is functorial in
$A$).

[Sch19] shows that these two maps are homotopic (Proposition 4.17). Let
$h_{\bullet}:F(A)\to F(A)$ be such a homotopy. Then, $2-[2]=dh+hd$.

Suppose $f\in\bigoplus_{j=1}^{n_{i}}C(A^{r{i,j}}\times S,\mathbb{R})$
satisfies $df=0$. Then,

$\displaystyle
2f-[2]^{*}(f)=d(h_{i-1}^{*}(f))+h_{i}^{*}(d(f))=d(h_{i-1}^{*}(f))\implies
f=\frac{1}{2}[2]^{*}(f)+d(\frac{1}{2}h_{i-1}^{*}(f))$

Inductively we obtain that

$\displaystyle
f=\frac{1}{2^{n}}[2^{n}]^{*}(f)+d\left(\sum_{k=1}^{n}\frac{1}{2^{k}}h_{i-1}^{*}([2^{n-1}]^{*}(f))\right)$

$[2^{n}]^{*}(f)$ is the map induced on $f$ when $A$ is multiplied by $2$. This
does not change the image so it is bounded with $\|[2^{n}]^{*}(f)\|\leq\|f\|$.
Also, $h_{i-1}^{*}$ is a map of Banach spaces so it must be bounded. Hence,
the sum converges and we can write

$\displaystyle
f=d\left(\sum_{k=1}^{\infty}\frac{1}{2^{k}}h_{i-1}^{*}([2^{n-1}]^{*}(f))\right)$

so the complex is exact. ∎

###  4.3 Locally compact abelian groups

We wil focus on a special class of topological groups, since they have a set
structure that simplifies calculations.

######  Definition 4.6.

A topological space is _locally compact_ if every point is contained in a
neighbourhood that is contained in some compact set.

A _locally compact abelian group_ is a Hausdorff abelian group whose
underlying topology is locally compact.

The following results may be found in [HS07, Proposition 2.2].

######  Theorem 4.7 (Structure of LCA groups).

Let $A$ be a locally compact abelian group.

  1. 1.

$A\cong\mathbb{R}^{n}\times A^{\prime}$, where $A^{\prime}$ is an extension of
a discrete abelian group by a compact abelian group.

  2. 2.

The Pontryagin duality functor
$A\mapsto\mathbb{D}(A)=\operatorname{Hom}(A,\mathbb{R}/\mathbb{Z})$.
$\mathbb{D}(A)$ takes values in locally compact abelian groups and
$\mathbb{D}(\mathbb{D}(A))\cong A$.

  3. 3.

If $A$ is compact (resp. discrete) if and only if $\mathbb{D}(A)$ is discrete
(resp. compact).

###  4.4 RHom in locally compact abelian groups

The rest of this section will consist of simplifying the necessary
calculations for $R\underline{\operatorname{Hom}}$ of locally compact abelian
groups.

######  Proposition 4.8.

Let $M$ and $N$ be LCA groups. Then,
$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ reduces to
assuming $M,N=\mathbb{R}$ or an extension of a discrete abelian group by a
compact abelian group.

###### Proof.

By Theorem 4.7, $M\cong\mathbb{R}^{m}\times M^{\prime}$ where $M$ is an
extension of a discrete abelian group by a compact abelian group. Similar for
$N$. We are done now observing that $R\underline{\operatorname{Hom}}(-,-)$
commutes with finite products (= direct sums) in either argument. ∎

######  Proposition 4.9.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where $M$ or
$N$ is an extension of a discrete abelian group by a compact abelian group
reduces to $M$ or $N$ is compact or discrete.

###### Proof.

Suppose $M$ is an extension of a discrete abelian group by a compact abelian
group. Then, it fits in a short exact sequence

${0}$${A}$${M}$${B}$${0}$$\scriptstyle{f}$$\scriptstyle{g}$

where $A$ is compact and $B$ is discrete.

We are going to prove that the sequence

${0}$${\underline{A}}$${\underline{M}}$${\underline{B}}$${0}$

is exact. By 2.18 it is enough to show that $f$ is a closed map and that for
every compact subset $B^{\prime}\subseteq B$ there is a compact subset
$M^{\prime}\subseteq M$ such that $g(M^{\prime})\supseteq B^{\prime}$.

$f$ is a continuous map from a compact space to a Hausdorff space, so it must
be closed. On the other hand, since $B$ is discrete, its only compact subsets
are finite, which are covered by finite (and hence compact) subsets of $M$.

This short exact sequence of condensed abelian groups induces a distinguished
triangle

$\displaystyle\underline{A}[0]\to\underline{M}[0]\to\underline{B}[0]\to\underline{A}[1]$

$R\underline{\operatorname{Hom}}:D(\operatorname{Cond(Ab)})^{op}\times
D(\operatorname{Cond(Ab)})$ is a morphism of triangulated categories when
fixing either argument, so we see that making the calculation for discrete and
compact is enough and then we may just complete the triangle. Similar if $N$
is an extension of a discrete abelian group by a compact abelian group. ∎

######  Proposition 4.10.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where $M$ is
discrete reduces to $M=\mathbb{Z}$.

###### Proof.

A discrete group $M$ has a two-term free resolution

${0}$${\mathbb{Z}[Q]}$${\mathbb{Z}[M]}$${M}$${0}$

where $\mathbb{Z}[Q]$ is the kernel of the obvious map $\mathbb{Z}[M]\to M$.
Since they are all discrete groups, the map $\mathbb{Z}[Q]\to\mathbb{Z}[M]$ is
closed. Also, any compact subset of $M$ must be finite so it is covered by a
finite (and hence compact) subset of $\mathbb{Z}[M]$. We may now apply 2.18
and see that the condensed sequence is exact.

We can now write $\mathbb{Z}[Q]=\bigoplus_{Q}\mathbb{Z}$ as a sum. Then, by
2.19, $\underline{\mathbb{Z}[Q]}=\bigoplus_{Q}\underline{\mathbb{Z}}$ so

$\displaystyle
R\underline{\operatorname{Hom}}(\underline{\mathbb{Z}[Q]},\underline{N})=R\underline{\operatorname{Hom}}(\bigoplus_{Q}\underline{\mathbb{Z}},\underline{N})=\prod_{Q}R\underline{\operatorname{Hom}}(\underline{\mathbb{Z}},\underline{N})$

Similar for
$R\underline{\operatorname{Hom}}(\underline{\mathbb{Z}[M]},\underline{N})$, so
we may just complete the triangle to compute
$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$. ∎

######  Proposition 4.11.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where $M$ is
compact reduces to $M=\prod_{I}\underline{\mathbb{R}/\mathbb{Z}}$ for some set
$I$.

###### Proof.

We will use Pontryagin duality. Since $M$ is compact, by the structure result
of locally compact abelian groups, $\mathbb{D}(M)$ is discrete. This then
gives a two-term free resolution

${0}$${\mathbb{Z}[Q]}$${\mathbb{Z}[\mathbb{D}(M)]}$${\mathbb{D}(M)}$${0}$

Pontryagin duality is an exact functor (it is an equivalence of categories)
so, noting that $\mathbb{D}(\mathbb{Z})=\mathbb{R}/\mathbb{Z}$ and sums factor
as products, we have a short exact sequence

${0}$${M}$${\prod_{I}\mathbb{R}/\mathbb{Z}}$${\prod_{J}\mathbb{R}/\mathbb{Z}}$${0}$

for some sets $I,J$. Now we will prove that the condensed sequence is exact.

The first map is from a compact set to a Hausdorff space, so it is closed.
Finally, notice that a compact subset of $\prod_{J}\mathbb{R}/\mathbb{Z}$ is
closed so its preimage is a closed subset of $\prod_{I}\mathbb{R}/\mathbb{Z}$,
which must be compact. We may now apply 2.18.

Hence, the induced sequence is exact and we may again get a distinguished
triangle. ∎

######  Proposition 4.12.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where $N$ is
compact reduces to $N=\mathbb{R}/\mathbb{Z}$.

###### Proof.

Use the same reasoning with Pontryagin duality and that products commute with
$R\underline{\operatorname{Hom}}$ in the second argument. ∎

######  Proposition 4.13.

We have the following similar cases.

  1. 1.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where
$M=\mathbb{R}$ reduces to $M=\mathbb{Z}$ and $M=\mathbb{R}/\mathbb{Z}$.

  2. 2.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where
$N=\mathbb{R}/\mathbb{Z}$ reduces to $N=\mathbb{R}$ and $N$ discrete.

###### Proof.

Use the short exact sequence

${0}$${\underline{\mathbb{Z}}}$${\underline{\mathbb{R}}}$${\underline{\mathbb{R}/\mathbb{Z}}}$${0}$

∎

With the above, we can finally compute $R\underline{\operatorname{Hom}}$ for
all locally compact abelian groups.

######  Theorem 4.14.

$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$ where $M$ and
$N$ are locally compact abelian groups reduces to the cases
$M=\prod_{I}\mathbb{R}/\mathbb{Z}$ for some set $I$ or $M=\mathbb{Z}$ and
$N=\mathbb{R}$ or $N$ discrete.

Observe that all of these cases were dealt with in the previous section.

The following diagrams summarise the simplifications made for
$R\underline{\operatorname{Hom}}(\underline{M},\underline{N})$.

$M$ LCA${M=\mathbb{R}}$${\begin{matrix}\text{$M$ ext. of disc.}\\\ \text{by
compact}\end{matrix}}$${M=\mathbb{Z}}$${M=\mathbb{R}/\mathbb{Z}}$$M$ disc.$M$
compact${M=\mathbb{Z}}$${M=\prod_{I}\mathbb{R}/\mathbb{Z}}$4.84.84.134.94.94.104.11

$N$ LCA${N=\mathbb{R}}$${\begin{matrix}\text{$N$ ext. of disc.}\\\ \text{by
compact}\end{matrix}}$$N$ disc.$N$
compact${N=\mathbb{R}/\mathbb{Z}}$${N=\mathbb{R}}$$N$
disc.4.84.84.94.94.114.134.13

This leads to the following interesting corollary. By following through all
the computations in the previous lemmas one can show that higher
$\underline{\operatorname{Ext}}$ groups vanish (c.f. [Sch19]). This means that
in the following examples we will focus on the first two ext groups.

######  Corollary 4.15.

If $M$ and $N$ are locally compact abelian groups,

$\displaystyle\underline{\operatorname{Ext}}^{i}(\underline{M},\underline{N})=0$

for $i\geq 2$.

######  Remark 4.16.

This seems to suggest that condensed abelian groups might have dimension $2$.
However, we remark that this is not true.

By Theorem 2.31, for a compact Hausdorff space $S$,

$\displaystyle\operatorname{Ext}^{i}(\mathbb{Z}[\underline{S}],\underline{\mathbb{Z}})=H^{i}(S,\mathbb{Z})\cong
H^{i}_{\text{sheaf}}(S,\mathbb{Z})$

which does not always vanish for $i\geq 2$ so it is impossible that we have
projective dimension $2$.

###  4.5 Examples

######  Example 4.17.

With the above one can obtain rather unsurprisingly that

$\displaystyle
R\underline{\operatorname{Hom}}(\underline{\mathbb{R}},\underline{\mathbb{R}^{\text{disc}}})=0$

since
$R\underline{\operatorname{Hom}}(\underline{\mathbb{Z}},\underline{\mathbb{R}^{\text{disc}}})=\underline{\mathbb{R}^{\text{disc}}}$
and
$R\underline{\operatorname{Hom}}(\underline{\mathbb{R}/\mathbb{Z}},\underline{\mathbb{R}^{\text{disc}}})=\underline{\mathbb{R}^{\text{disc}}}[-1]$
so we may just complete the distinguished triangle induced from the exact
sequence $0\to\mathbb{Z}\to\mathbb{R}\to\mathbb{R}/\mathbb{Z}\to 0$.

######  Example 4.18.

One can also prove that

$\displaystyle\underline{\operatorname{Hom}}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}/\mathbb{Z}})=\underline{\hat{\mathbb{Q}}}$
$\displaystyle\underline{\operatorname{Ext}}^{1}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}/\mathbb{Z}})=0$

where $\hat{\mathbb{Q}}$ is the solenoid, the Pontryagin dual of
$\mathbb{Q}^{\text{disc}}$.

The first part is clear. We may now use again a two-term resolution of
condensed abelian groups. The long exact sequence on $\operatorname{Ext}$
groups gives

$\displaystyle
0\to\hat{\mathbb{Q}}\to\widehat{\mathbb{Z}[\mathbb{Q}]}\to\widehat{\mathbb{Z}[K]}\to
0$

All these groups are Pontryagin duals of discrete spaces so they are compact.
Hence, the first map is from a compact space to a Hausdorff space, so it is
closed. Any compact subset of $\widehat{\mathbb{Z}[K]}$ is closed so it has a
closed and therefore compact preimage in $\widehat{\mathbb{Z}[\mathbb{Q}]}$.
By 2.18 the condensed sequence is exact.

Using the long exact sequence for enriched $\operatorname{Ext}$ groups,

$\displaystyle
0\to\underline{\hat{\mathbb{Q}}}\to\underline{\widehat{\mathbb{Z}[\mathbb{Q}]}}\to\underline{\widehat{\mathbb{Z}[K]}}\to\underline{\operatorname{Ext}}^{1}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}/\mathbb{Z}})\to
0$

But due to the condensed short exact sequence,
$\underline{\widehat{\mathbb{Z}[\mathbb{Q}]}}\to\underline{\widehat{\mathbb{Z}[K]}}$
is a surjection and

$\displaystyle\underline{\operatorname{Ext}}^{1}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}/\mathbb{Z}})=0$

######  Example 4.19.

Suppose we want to calculate
$R\underline{\operatorname{Hom}}(\underline{\mathbb{R}^{\text{disc}}},\underline{\mathbb{R}})$.

By Zorn’s lemma, choose a basis $\mathcal{B}$ of $\mathbb{R}$ over
$\mathbb{Q}$. [Neu28] gives explicit examples of algebraically independent
subsets of $\mathbb{R}$ over $\mathbb{Q}$ of cardinality $2^{\aleph_{0}}$ and
$2^{\aleph_{0}}$ is also a trivial upper bound for $|\mathcal{B}|$ so
$|\mathcal{B}|=2^{\aleph_{0}}$. Then,

$\displaystyle
R\underline{\operatorname{Hom}}(\underline{\mathbb{R}^{\text{disc}}},\underline{\mathbb{R}})=R\underline{\operatorname{Hom}}(\bigoplus_{\mathcal{B}}\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})=\prod_{\mathcal{B}}R\underline{\operatorname{Hom}}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})=R\underline{\operatorname{Hom}}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})^{2^{\aleph_{0}}}$

We first have that

$\displaystyle\underline{\operatorname{Ext}}^{0}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})=\underline{\operatorname{Hom}}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})=\underline{\operatorname{Hom}(\mathbb{Q}^{\text{disc}},\mathbb{R})}$

Any function $\mathbb{Q}^{\text{disc}}\to\mathbb{R}$ is continuous so we only
need to look at group homomorphisms, which are functions satisfying
$f(x+y)=f(x)+f(y)$ for all $x,y\in\mathbb{Q}^{\text{disc}}$.

This is exactly Cauchy’s functional equation, whose only solutions are of the
form $f(x)=\alpha x$ for some $\alpha\in\mathbb{R}$ [MK08, Chapter 5], so the
underlying set is $\mathbb{R}$. Now we must compute the topology.

We claim that the compact-open topology gives $\mathbb{R}$ its natural
topology. There is a subbasis consisting of the following sets, for compact
$K\subseteq\mathbb{Q}^{\text{disc}}$ and open $U\subseteq\mathbb{R}$

$\displaystyle V(K,U)=\\{\alpha\in\mathbb{R}|\alpha K\subseteq U\\}$

Clearly, $V(\\{1\\},(a,b))=(a,b)$. Open intervals generate the Euclidean
topology on $\mathbb{R}$, so the compact-open topology is no coarser than the
natural one.

Only finite sets of $\mathbb{Q}^{\text{disc}}$ are compact. We write $V(K,U)$
as an intersection of finitely many open sets.

$\displaystyle V(K,U)=\bigcap_{k\in K}V(\\{k\\},U)$

Also, $U$ is open so it can be written as $U=\bigcup_{i\in I}(a_{i},b_{i})$
and clearly $V(K,U)=\bigcup_{i\in I}V(K,(a_{i},b_{i}))$. Finally, notice that
$V(\\{k\\},(a_{i},b_{i})))=(a_{i}/k,b_{i}/k)$. Hence,

$\displaystyle V(K,U)=\bigcap_{k\in K}\bigcup_{i\in I}(a_{i}/k,b_{i}/k)$

which is clearly open under the natural topology of $\mathbb{R}$ (we omit the
case $k=0$ because $V(\\{0\\},U)=\emptyset$ or $\mathbb{R}$ so it follows
trivially). This shows that the compact-open topology gives $R$ its natural
topology.

Hence,

$\displaystyle\underline{\operatorname{Ext}}^{0}(\underline{\mathbb{R}^{\text{disc}}},\underline{\mathbb{R}})=\underline{\mathbb{R}}^{2^{\aleph_{0}}}$

The next enriched extension group is not known by the author. By our methods
described, one would use a two-term free resolution

$\displaystyle
0\to\mathbb{Z}[K]\to\mathbb{Z}[\mathbb{Q}]\to\mathbb{Q}^{\text{disc}}\to 0$

to obtain a distinguished triangle. The main problem is the behaviour of the
map $\mathbb{Z}[K]\to\mathbb{Z}[\mathbb{Q}]$, since it is difficult to obtain
an explicit description of the kernel of
$\mathbb{Z}[\mathbb{Q}]\to\mathbb{Q}$.

It is conjectured by the author that
$\underline{\operatorname{Ext}}^{1}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})=0$,
by comparison with the category of abelian groups. Additionally, due to the
fully faithful embedding from the category of LCA groups [Sch19, Corollary
4.9], we know that
$\underline{\operatorname{Ext}}^{1}(\underline{\mathbb{Q}^{\text{disc}}},\underline{\mathbb{R}})(*)=0$
by the known result there.

##  5 A prismatic construction of the real numbers

This section constructs the real numbers as a quotient of condensed rings. The
motivation for such a construction lies in the theory of $p$-liquid vector
spaces, a full subcategory of condensed vector spaces stable under all limits,
colimits and extensions [Sch]. This is the functional analysis analogue to the
solid theory for condensed abelian groups [Sch19].

The building blocks for this theory are condensed objects of the form
$\mathcal{M}_{p}(S)$, which can be thought of as $p$-measures on a profinite
set $S$. To obtain a category similar to solid abelian groups, we would want
these to be projective. This is not quite the case, but one can obtain the
close result that $\operatorname{Ext}^{i}(\mathcal{M}_{p}(S),V)=0$ for $i>0$
and any $p^{\prime}$-Banach vector space with $0<p<p^{\prime}\leq 1$.

To simplify the problem, Scholze and Clausen build $\mathbb{R}$ as a quotient
of discrete spaces. This section will only explore this quotient, without any
further analysis of the liquid theory.

###  5.1 The main result

We first start with the condensed ring from which we will construct the real
numbers.

######  Definition 5.1.

For $0<r<1$ and $c>0$ we define the condensed set
$\underline{\mathbb{Z}((T))_{r,\leq c}}$ with $S$-valued points

$\displaystyle\underline{\mathbb{Z}((T))_{r,\leq
c}}(S):=\\{\sum_{n\gg-\infty}a_{n}T^{n}|a_{n}\in C(S,\mathbb{Z}),\forall s\in
S\;\sum|a_{n}(s)|r^{n}\leq c\\}$

This allows the definition of the condensed ring

$\displaystyle\mathbb{Z}((T))_{r}=\bigcup_{c>0}\mathbb{Z}((T))_{r,\leq c}$

######  Remark 5.2.

One would need to check that $\underline{\mathbb{Z}((T))_{r,\leq c}}$ is a
condensed set. Rather than carrying out a construction with Kan extensions, we
will prove in the next subsection that it is the condensation of a T1
topological space.

Also, observe that one obtains a ring at each section after the union, so the
resulting condensed set $\mathbb{Z}((T))_{r}$ is actually a condensed ring.

We now have the following result.

######  Theorem 5.3.

Let $0<r^{\prime}<r<1$ and consider the map

$\displaystyle\theta_{r^{\prime}}:\mathbb{Z}((T))_{r}$
$\displaystyle\to\underline{\mathbb{R}}$ $\displaystyle\sum a_{n}T^{n}$
$\displaystyle\mapsto\sum a_{n}(r^{\prime})^{n}$

This is a surjection and the kernel at every section is a principal ideal
generated by a non-zero divisor.

######  Remark 5.4.

This result looks initially rather unintuive but it is best understood in
terms of decimal expansion. Indeed, set $r^{\prime}=\frac{1}{10}$. Then, this
is equivalent to writing numbers in the form

$\displaystyle\sum_{n\gg-\infty}a_{n}10^{-n}$

which is precisely decimal expansion. In this case, we will prove later that
the kernel is generated at every section by the polynomial $10T-1$, which
would be the expected behaviour.

We highlight the similarity with the following number-theoretic result.

######  Theorem 5.5.

Let $K$ be a complete discretely valued extension of $\mathbb{Q}_{p}$ with
perfect residue field $k$. Then, the ring of integers $\mathcal{O}_{K}$ can be
written as a quotient of $W(k)[[T]]$ by a principal ideal generated by an
Eisenstein polynomial $f$.

This is the reason for calling this a prismatic construction. By Scholze’s and
Bhatt’s recent work in prismatic cohomology, the pair $(W(k)[[T]],(f))$ is a
prism [BS19].

###  5.2 A helpful ring

We first look at a non-condensed version of $\mathbb{Z}((T))_{r,\leq c}$.

######  Definition 5.6.

For $0<r<1$ and $c>0$ define the topological space

$\displaystyle\mathbb{Z}((T))_{r,\leq
c}:=\\{\sum_{n\gg-\infty}a_{n}T^{n}|a_{n}\in\mathbb{Z},\sum|a_{n}|r^{n}\leq
c\\}$

With the topology induced by the $T$-adic norm

$\displaystyle\lVert\sum a_{n}T^{n}\rVert=\delta^{v_{T}(f)}$

for some fixed $\delta\in(0,1)$ where

$\displaystyle v_{T}(f)=\sup\\{n\in\mathbb{Z}:a_{n}=0\\}$

The following lemma shows that our notation is consistent.

######  Lemma 5.7.

The condensed set $\underline{\mathbb{Z}((T))_{r,\leq c}}$ is the condensation
of $\mathbb{Z}((T))_{r,\leq c}$.

###### Proof.

Fix extremally disconnected $S$. We must prove that

$\displaystyle C(S,\mathbb{Z}((T))_{r,\leq
c})=\underline{\mathbb{Z}((T))_{r,\leq
c}}(S)=\\{\sum_{n\gg-\infty}a_{n}T^{n}|a_{n}\in C(S,\mathbb{Z}),\forall s\in
S\;\sum|a_{n}(s)|r^{n}\leq c\\}$

Let $F\in C(S,\mathbb{Z}((T))_{r,\leq c})$. Then, for any $s\in S$ we can
write

$\displaystyle F(s)=\sum_{n\in\mathbb{Z}}a_{n}(s)T^{n}$

for some $a_{n}:S\to\mathbb{Z}$ not necessarily continuous, where for all
$s\in S$, $\sum|a_{n}(s)|r^{n}\leq c$. The space $S$ is compact, so its image
is bounded. Hence, there is $k\in\mathbb{Z}$ such that $a_{n}=0$ for all
$n<k$. It is clear that for all $s\in S$, $\sum|a_{n}|r^{n}\leq c$.

We must now prove the $a_{n}$ are continuous. We will show that for any
$A\subseteq S$, $a_{n}(\overline{A})\subseteq\overline{a_{n}(A)}$. Since
$\mathbb{Z}$ has the discrete topology, this amounts to proving
$a_{n}(\overline{A})\subseteq a_{n}(A)$.

Let $s\in\overline{A}$. Then, $F$ is continuous so $F(s)\in\overline{F(A)}$
and for any $\varepsilon>0$, there exists $s_{\varepsilon}\in A$ such that
$\lVert F(s)-F(s_{\varepsilon})\rVert<\varepsilon$. Setting
$\varepsilon<\delta^{n}$ we see that the coefficients of $T^{k}$ must coincide
for $k\leq n$, so $a_{n}(s)=a_{n}(s_{\varepsilon})\in a_{n}(A)$. Hence,
$a_{n}$ is continuous.

To prove the other direction, one must show that if $a_{n}\in C(S,\mathbb{Z})$
and for all $s\in S$, $\sum_{n\gg-\infty}|a_{n}(s)|r^{n}\leq c$, then

$\displaystyle F=\sum_{n\gg-\infty}a_{n}T^{n}\in C(S,\mathbb{Z}((T))_{r,\leq
c})$

Let $A\subseteq S$ and $s\in\overline{A}$. Each $a_{n}$ partitions $S$ into
finitely many clopen subsets, so by taking intersections, for any
$N\in\mathbb{N}$ there are clopen sets $U_{1},\dots,U_{k}$ such that $a_{n}$
is constant on $U_{i}$ for $n\leq N$.

$s\in U_{i}$ for some $i$. Since $U_{i}$ is open and $s\in\overline{A}$,
$U_{i}\cap A\neq\emptyset$. Let $s^{\prime}\in U_{i}\cap A$. $a_{n}$ is
constant on the $U_{i}$ for $n\leq N$ so

$\displaystyle\lVert F(s)-F(s^{\prime})\rVert\leq\delta^{N}\to 0$

as $N\to\infty$ so $F(s)$ is indeed a limit point. ∎

While $\underline{\mathbb{Z}((T))_{r}}$ is not the condensation of a compact
space, it is built by compact sets in a similar way to the real numbers.

######  Lemma 5.8.

$\mathbb{Z}((T))_{r,\leq c}$ is compact.

###### Proof.

We will prove the stronger result that it is a profinite, which is equivalent
to being a limit of finite discrete spaces [Sta21, Tag 08ZY].

Firstly, $\sum|a_{i}|r^{i+k}=r^{k}\sum|a_{i}|r^{i+k}$ so multiplying by
$T^{k}$ is an isomorphism $\mathbb{Z}((T))_{r,\leq
c}\to\mathbb{Z}((T))_{r,r^{k}c}$. Hence, multiplying by a high enough power of
$T$, we may assume that $c<1$ (remember that $r<1$).

Then,

$\displaystyle\mathbb{Z}((T))_{r,\leq
c}\subseteq\mathbb{Z}[[T]]=\lim_{\longleftarrow}\prod_{n=0}^{m}\mathbb{Z}\cdot
T^{n}$

where the inverse limit is of discrete topological spaces, and the right-hand
space has the $T$-adic topology.

Define

$\displaystyle\mathbb{Z}((T))_{r,\leq
c}^{m}=\\{\sum_{n=0}^{m}a_{n}T^{n}|a_{n}\in\mathbb{Z},\sum_{n=0}^{m}|a_{n}|r^{n}\leq
c\\}\subseteq\prod_{n=0}^{m}\mathbb{Z}\cdot T^{n}$

Since $\sum|a_{n}|r^{n}\leq c$ we have finite choices of $a_{n}$ so the sets
$\mathbb{Z}((T))_{r,\leq c}^{m}$ are finite.

In an analogous way to $\mathbb{Z}[[T]]$ we can write $\mathbb{Z}((T))_{r,\leq
c}$ as an inverse limit of the $\mathbb{Z}((T))_{r,\leq c}^{m}$ and it
inherits the $T$-adic topology. ∎

To understand the condensed ring $\mathbb{Z}((T))_{r}$ we first study the
following subring of power series.

######  Definition 5.9.

Let $0<r<1$. Define the ring

$\displaystyle\mathbb{Z}((T))_{>r}=\\{\sum_{n\gg-\infty}a_{n}T^{n}|a_{n}\in\mathbb{Z},\;\exists
r^{\prime}>r:|a_{n}|(r^{\prime})^{n}\to 0\\}$

######  Remark 5.10.

The condition $|a_{n}|(r^{\prime})^{n}\to 0$ tells us the following.
$|a_{n}|(r^{\prime})^{n}$ must be bounded as a sequence so for any
$r<s<r^{\prime}$,

$\displaystyle\lVert\sum_{n\gg-\infty}a_{n}s^{n}\rVert\leq\sum_{n\gg-\infty}(|a_{n}|(r^{\prime})^{n})\left(\frac{s}{r^{\prime}}\right)^{n}<\infty$

by comparison with the geometric series, so the radius of convergence is $\geq
r^{\prime}$. Hence, $\mathbb{Z}((T))_{>r}$ are all holomorphic functions on
some neighbourhood on $B(0,r)\setminus\\{0\\}$.

Notice that naturally
$\mathbb{Z}((T))_{>r}\subseteq\underline{\mathbb{Z}((T))_{r}}(*)$.

This ring is extensively studied in [Sch] and [Har84]. We quote the following
result.

######  Theorem 5.11.

Let $0<r<1$. Then, $\mathbb{Z}((T))_{>r}$ is a principal ideal domain. The
non-zero prime ideals are the following.

  1. 1.

For any $x\in\mathbb{C}^{*}$ with $|x|\leq r$, the kernel of the map

$\displaystyle\mathbb{Z}((T))_{>r}\to\mathbb{C}:\sum a_{n}T^{n}\mapsto\sum
a_{n}x^{n}$

generated by some $f\in 1+T\mathbb{Z}[[T]]$ whose only zeroes on
$\overline{B}(0,r)$ are at $x$ and $\overline{x}$ with multiplicity $1$. This
map is surjective onto $\mathbb{R}$ when $x\in\mathbb{R}$.

  2. 2.

For any prime number $p$, the ideal $(p)$.

  3. 3.

For any prime number $p$ and any topologically nilpotent unit
$x\in\overline{\mathbb{Q}_{p}}$, the kernel of the map

$\displaystyle\mathbb{Z}((T))\to\overline{\mathbb{Q}_{p}}:\sum
a_{n}T^{n}\mapsto a_{n}x^{n}$

###  5.3 Proof of the main result

We will need a few lemmas in order to conclude the proof. The main proof
strategy will be restricting ourselves to the compact sets
$\mathbb{Z}((T))_{r,\leq c}$ and using the fact that extremally disconnected
spaces are projective in the category of compact Hausdorff spaces.

######  Lemma 5.12.

Let $0<r<1$ and let $r^{\prime}<r$. Then, the map

$\displaystyle F:\mathbb{Z}((T))_{r,\leq
c}\to\mathbb{R},\;\sum_{n\gg-\infty}a_{n}T^{n}\mapsto\sum_{n\gg-\infty}a_{n}(r^{\prime})^{n}$

is continuous.

###### Proof.

For $f=\sum_{n\gg-\infty}a_{n}T^{n}\in\mathbb{Z}((T))_{r,\leq c}$ we must have
$a_{n}\leq c\cdot r^{-n}$ for all $n$. If $\lVert f-g\rVert<\delta^{N}$, the
first $N$ coefficients match so

$\displaystyle\lVert F(f)-F(g)\rVert\leq
2\sum_{n>N}cr^{-n}(r^{\prime})^{n}=2c\frac{\left(\frac{r^{\prime}}{r}\right)^{N}}{1-\frac{r^{\prime}}{r}}\to
0$

as $N\to\infty$ so the function is continuous. ∎

######  Lemma 5.13.

Let $K\subseteq\mathbb{R}$ be compact. Then, there is some $c>0$ such that

$\displaystyle K\subseteq\theta_{r^{\prime}}(\mathbb{Z}((T))_{r,\leq c})$

where $\theta_{r^{\prime}}:\mathbb{Z}((T))_{r,\leq c}\to\mathbb{R}$ is the map
in Theorem 5.3 evaluated at the point.

###### Proof.

Without loss of generality, assume $K=[-M,M]$ since any compact subset of
$\mathbb{R}$ is contained in one such set.

Fix $x\in K\setminus\\{0\\}$, and pick $n$ minimal so that
$(r^{\prime})^{n}\leq|x|$. This minimum is well-defined since
$(r^{\prime})^{n}\to 0$ as $n\to\infty$ and $(r^{\prime})^{n}\to\infty$ as
$n\to-\infty$. Also, there is a lower bound on $n$ that is uniform for all
$x\in K$ (precisely, $n\geq\log M/\log r^{\prime}$).

Pick an integer $a_{n}\in\mathbb{Z}$ such that

$\displaystyle\left|\frac{x}{(r^{\prime})^{n}}-a_{n}\right|<1$

and let $x^{\prime}=x-a_{n}(r^{\prime})^{n}$, so by our choice of $a_{n}$,
$|x^{\prime}|<(r^{\prime})^{n}\leq|x|$ and $x^{\prime}\in K$.

Due to the choice of minimal $n$,

$\displaystyle(r^{\prime})^{n}\leq|x|<(r^{\prime})^{n-1}$

So we must have

$\displaystyle|a_{n}|=\left|\frac{x}{(r^{\prime})^{n}}-\frac{x^{\prime}}{(r^{\prime})^{n}}\right|\leq\frac{|x|}{(r^{\prime})^{n}}+\frac{|x^{\prime}|}{(r^{\prime})^{n}}<\frac{1}{r^{\prime}}+1$

We now repeat the process with $x^{\prime}$, choosing an $n^{\prime}$ minimal
so that $(r^{\prime})^{n^{\prime}}\leq|x^{\prime}|$. By our choice of
$x^{\prime}$ we must have $n^{\prime}>n$, and by the same reasoning
$|a_{n^{\prime}}|<1+\frac{1}{r^{\prime}}$. Hence,

$\displaystyle x=\sum_{n\gg-\infty}a_{n}(r^{\prime})^{n}$

where $|a_{n}|<1+1/r^{\prime}$ for all $n$. Then, $x=\theta_{r^{\prime}}(\sum
a_{n}T^{n})$. Finally,

$\displaystyle\sum|a_{n}|r^{n}\leq(1+\frac{1}{r^{\prime}})\sum_{n\geq\log
M/\log r^{\prime}}r^{n}$

which is a bound only dependent on $M$, so
$\theta_{r^{\prime}}(\mathbb{Z}((T))_{r,\leq c})\supseteq K$ for some $c>0$. ∎

######  Lemma 5.14.

The kernel of $\theta_{r^{\prime}}$ as in Theorem 5.3 is generated by some
$f_{r^{\prime}}\in\mathbb{Z}((T))_{>r}$ at every section.

###### Proof.

Consider the map

$\displaystyle\mathbb{Z}((T))_{>r}\to\mathbb{R}:\sum a_{n}T^{n}\mapsto\sum
a_{n}(r^{\prime})^{n}$

By Theorem 5.11, the kernel is generated by some
$f_{r^{\prime}}\in\mathbb{Z}((T))_{>r}$, where $f\in 1+T\mathbb{Z}((T))$ and
only vanishes at $r^{\prime}$ within $\overline{B}(0,r)$.

We claim that the kernel of $\theta_{r^{\prime}}$ is generated by
$f_{r^{\prime}}$ at every section (since it has constant functions on the
coefficients, $f_{r^{\prime}}\in\mathbb{Z}((T))_{r}(S)$ for every $S$).

Let $S$ be extremally disconnected, and take
$g=\sum_{n\gg-\infty}a_{n}T^{n}\in\mathbb{Z}((T))_{r}(S)$ such that
$\theta_{r^{\prime}}(g)=0$. Now write $f_{r^{\prime}}=1+\sum_{n\geq
1}b_{n}T^{n}$ and notice that near $T=0$ we can write

$\displaystyle\frac{1}{f_{r^{\prime}}}=\frac{1}{1+\sum_{n\geq
1}b_{n}T^{n}}=1+\sum_{k=0}^{\infty}(-1)^{k}\left(\sum_{n\geq
1}b_{n}T^{n}\right)^{k}\in\mathbb{Z}[[T]]$

This is well-defined in a neighbourhood of $\overline{B}(0,r)$ except at
$T=r^{\prime}$, where we have a simple pole. Also, all the coefficients of the
series are (constant) continuous functions $S\to\mathbb{Z}$.

Then, $g\cdot f^{-1}_{r^{\prime}}$ has integer coefficients and a removable
singularity at $T=r^{\prime}$. Write $|g|_{r,s}$ for $\sum|a_{n}(s)|r^{n}$ so
that $|g|_{r,s}\leq c$ for some $c>0$ and all $s\in S$. Note that
$f\in\mathbb{Z}((T))_{>r}$ so $f^{-1}_{r^{\prime}}$ is defined in an annulus
centered at $0$ that contains $r$, and by absolute convergence
$|f^{-1}_{r^{\prime}}|_{r,s}\leq c^{\prime}$ for some $c^{\prime}>0$ and all
$s\in S$ ($f_{r^{\prime}}$ is constant as a function on $S$). Hence, $|g\cdot
f^{-1}_{r^{\prime}}|_{r,s}\leq|g|_{r,s}\cdot|f^{-1}_{r^{\prime}}|_{r,s}\leq
c\cdot c^{\prime}$ for all $s\in S$. Therefore, $g\cdot
f^{-1}_{r^{\prime}}\in\mathbb{Z}((T))_{r}(S)$.

It follows that $g\in(f_{r^{\prime}})$.

∎

######  Lemma 5.15.

The function
$\theta_{r^{\prime}}:\mathbb{Z}((T))_{r}\to\underline{\mathbb{R}}$ is an
epimorphism.

###### Proof.

Evaluating at every section, we must check that there is a surjection

$\displaystyle\mathbb{Z}((T))_{r}(S)\to C(S,\mathbb{R})$

Let $f\in C(S,\mathbb{R})$. Then, $f(S)$ is compact so by Lemma 5.13 there is
some $c>0$ such that

$\displaystyle\theta_{r^{\prime}}(\mathbb{Z}((T))_{r,\leq c})\supseteq f(S)$

$f(S)\subseteq\mathbb{R}$ is compact, so it is closed. By Lemma 5.8,
$\mathbb{Z}((T))_{r,\leq c}$ is compact and by Lemma 5.12, the map
$\theta_{r^{\prime}}$ is continuous when restricted to
$\mathbb{Z}((T))_{r,\leq c}$. Hence,
$\theta_{r^{\prime}}^{-1}(f(S))\cap\mathbb{Z}((T))_{r,\leq c}$ is a closed
subset of a compact space, so it is compact.

${S}$${\theta_{r^{\prime}}^{-1}(f(S))\cap\mathbb{Z}((T))_{r,\leq
c}}$${f(S)}$$\scriptstyle{f}$$\scriptstyle{\theta_{r^{\prime}}}$

We have the following solid diagram of compact Hausdorff spaces. Since
extremally disconnected sets are projective in the category of compact
Hausdorff spaces, this factors through some continuous map
$g:S\to\theta_{r^{\prime}}^{-1}(f(S))\cap\mathbb{Z}((T))_{r,\leq
c}\subseteq\mathbb{Z}((T))_{r,\leq c}$.

By Lemma 5.7, $g\in\underline{\mathbb{Z}((T))_{r,\leq
c}}(S)\subseteq\mathbb{Z}((T))_{r}(S)$ so we have $\theta_{r^{\prime}}(g)=f$.
The map is surjective at every section, so it is an epimorphism.

∎

###### Proof.

(of Theorem 5.3) We will show that

$\displaystyle
0\to\mathbb{Z}((T))_{r}\xrightarrow{f_{r^{\prime}}}\mathbb{Z}((T))_{r}\xrightarrow{\theta_{r^{\prime}}}\underline{\mathbb{R}}\to
0$

is a short exact sequence of condensed abelian groups. To prove the first
injection, we must show that for profinite $S$,
$f_{r^{\prime}}\in\mathbb{Z}((T))_{r}(S)$ is not a zero divisor. Evaluating at
$s\in S$, $f_{r^{\prime}}$ does not change because it only has constant
functions as its coefficients.

Furthermore, $\mathbb{Z}((T))_{r}(S)(s)$ is an integral domain, so if
$f(s)g(s)=0$ then $g(s)=0$. Then, if $fg=0$, $g(s)=0$ for all $S$ and $g=0$.

By Lemma 5.14,
$\ker(\theta_{r^{\prime}})=f_{r^{\prime}}\cdot\mathbb{Z}((T))_{r}$. Finally,
by Lemma 5.15, $\theta_{r^{\prime}}$ is an epimorphism so this is a short
exact sequence. ∎

######  Example 5.16.

We now return to the mentioned example, setting $r^{\prime}=\frac{1}{10}$.
Since $f_{r^{\prime}}\in 1+T\mathbb{Z}[[T]]$, we have $\deg f_{r^{\prime}}\geq
1$. Evaluating at the point,

$\displaystyle\mathbb{Z}((T))_{r}(*)=\\{\sum_{n\gg-\infty}a_{n}T^{n}|\sum
a_{n}r^{n}<\infty\\}$

Within this ring, we must have $f_{r^{\prime}}|10T-1$ since it is in the
kernel. Write $f_{r^{\prime}}g=10T-1$. Since $f_{r^{\prime}}$ does not vanish
at $0$ and $10T-1$ does not have a pole at $0$, $g\in\mathbb{Z}[[T]]$. But
then for degree reasons it is clear $\deg g=0$ so $f_{r^{\prime}}=\pm(10T-1)$.

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