# When is a complex of Banach spaces exact as condensed abelian groups?

Lectures on Condensed Mathematics, Theorem 3.3 says that for any compact Hausdorff space $$S$$, the cohomology $$H_{\mathrm{cond}}^i(S,\mathbb R)=0$$ for $$i>0$$ and $$H_{\mathrm{cond}}^0(S,\mathbb R)=C(S,\mathbb R)$$ the space of continuous functions on $$S$$. Furthermore, a "quantitative" version is given. It is not immediately clear to me any application of this estimation (I don't know how much it depends on the choice of hypercover, either). Later, I realize that it might be related to the following question:

Let $$V'\xrightarrow fV\xrightarrow gV''$$ be a complex of real Banach spaces (i.e. $$g\circ f=0$$). When is it exact at $$V$$, that is to say, the induced map $$\operatorname{im}(f)\to\ker(g)$$, taken in the abelian category of condensed abelian groups, is an equivalence? This should be equivalent to the condition that, for any extremally disconnected set $$T$$, the induced complex $$C(T,V')\to C(T,V)\to C(T,V'')$$ is exact at $$C(T,V)$$ (as abelian groups).

In particular, the natural counterpart of the "quantitative" version of exactness might look like this: for any "cycle" $$v\in V$$ (i.e. $$g(v)=0$$) and for all $$\varepsilon>0$$, there exists $$v'\in V'$$ such that $$f(v')=v$$ and $$\lVert v'\rVert\le(1+\varepsilon)\lVert v\rVert$$. Does this imply the exactness of $$V'\to V\to V''$$ as condensed abelian groups? I failed to construct a proof for this simple proposition. When the index category is $$\mathbb N$$, given a pro-finite set $$T=\lim_{i\in\mathbb N}T_i$$ such that each $$T_i$$ is finite and the transition maps are surjective, and a continuous map $$h\colon T\to V$$ such that $$g\circ h=0$$, since $$T$$ is compact, $$h$$ is uniformly continuous. It looks like then we can use the quantitative version to construct inductively a continuous map $$q\colon T\to V'$$ such that $$f\circ q=h$$. I don't know how to proceed for general cofiltered index category. Zorn's lemma might be necessary.

If I am not mistaken, the condensed abelian group $$C(S,\mathbb R)$$ is also "equivalent" to the internal Hom: for any extremally disconnected set $$T$$, we have $$\underline{C(S,\mathbb R)}(T)\cong C(T,C(S,\mathbb R))\cong C(S\times T,\mathbb R)\cong\underline{\operatorname{Hom}}_{\operatorname{Cond}(\operatorname{Ab})}(\mathbb Z[S],\mathbb R)(T)$$.

Update. Sorry for my silly question. I realize that this is essentially characterizing maps of Banach spaces being epimorphisms in the abelian category of condensed abelian groups. And my previous argument does not seem to be related to the quantitative statement in Scholze's notes. Indeed, the open mapping theorem guarentees that such a constant bound always exists since $$\operatorname{im}(f)=\ker(g)$$ is Banach.

Thanks for the nice question. As you say in the update, the issue is basically the following: suppose given a map of Banach spaces $$f:V\rightarrow W$$. When is the induced map of condensed $$\mathbb{R}$$-modules $$\underline{V}\rightarrow\underline{W}$$ an epimorphism? This is not obvious to answer from first principles.

Nonetheless the answer is simple: $$\underline{f}:\underline{V}\rightarrow\underline{W}$$ is an epimorphism if and only if $$f$$ is surjective. Moreover, this holds also for Frechet spaces $$V$$ and $$W$$.

For the proof, it suffices to see that if $$f$$ is surjective, then for every compact subset $$K\subset W$$ there is a compact subset $$L\subset V$$ with $$f(L)=K$$. Probably this is a standard fact in functional analysis. To prove it I like to use the nice lemma of Grothendieck that every compact subset of a Frechet space is contained in the closure of the convex hull of a null-sequence. Moreover such closed convex hulls of null sequences are themselves compact. This reduces us to showing that every nullsequence in $$W$$ lifts to a nullsequence in $$V$$. But that is a consequence of the open mapping theorem, as you note.

The upshot, combining again with the open mapping theorem, is that a complex of Frechet spaces is exact when viewed as a complex of condensed $$\mathbb{R}$$-modules if and only if it is exact on underlying real vector spaces, and moreover in such a case it is strictly exact in the usual sense of topological vector spaces, so the subspace topology on ker(d) equals the quotient topology on im(d).

You are also correct that the condensed $$\mathbb{R}$$-module associated to the Banach space $$C(S,V)$$ with sup norm is the same as the internal Hom in condensed $$R$$-modules from $$\mathbb{R}[S]$$ to $$V$$. (Here $$S$$ is compact Hausdorff and $$V$$ is Banach.)

• I want to add a remark (after reading other MO posts these days): any surjective continuous linear map of Banach spaces admets a continuous (non-linear) section, called Bartle-Graves theorem. Here no compactness manifestly appears. In particular, the "standard fact" follows since the continuous image of a compact set is compact.
– user20948
Feb 11, 2021 at 20:07
• Oh, interesting. So that means that $C(K,V) \rightarrow C(K,W)$ is surjective for all topological spaces $K$, not just extremally disconnected $K$. If $K$ is profinite I think this can also be proved directly from the open mapping theorem as follows, and this is probably a proof more in the spirit of your question: note that $C(K,V)$ is the completion of the normed vector space $C_{lc}(K,V)$ of locally constant functions w.r.t the sup norm. Also, from the open mapping theorem, $C_{lc}(K,V)\rightarrow C_{lc}(K,W)$ is better than surjective: we can also uniformly bound the norm of a preimage. Feb 22, 2021 at 15:13
• Now use the general lemma that if $A\rightarrow B$ is such a "uniformly" surjective map of normed vector spaces, then the induced map of completions is also surjective. Proof: represent an element in the completion of B as $\sum b_n$ where the norm of $b_n$ decreases rapidly to $0$ (always possibly by passing to a subsequence in one's Cauchy sequence), then lift the terms individually. Feb 22, 2021 at 15:19