I have come upon a gap in lecture notes I am working through and would appreciate some help. I'm considering the following:

Let $H_\ast$ denote singular homology. Given a filtration

$$ \emptyset =: X_{-1} \subset X_1 \subset \dots \subset X = \bigcup_{i\ge -1} X_i $$ of topological spaces, where $X$ carries the colimit topology w.r.t. the family $\left(X_i\right)_{i\ge-1}$, such that for $n\ge 0$ and $i \ge 1$, $$ H_n\left(X_{n+1}\right) \underset{H_n(X_{n+1}\subset X_{n+2})}{\overset\cong\longrightarrow} H_n \left(X_{n+2}\right) \underset{H_n(X_{n+2}\subset X_{n+3})}{\overset \cong \longrightarrow} \dots \underset{H_n(X_{n+i-1}\subset X_{n+i})}{\overset\cong\longrightarrow} H_n \left(X_{n+i}\right)\,, $$ how can one infer that $$ H_n\left(X_{n+i}\right) \cong H_n(X)\quad \text{?} $$

I know that there is nothing to show if $X$ has finite dimension, so I'm interested in the infinite case.

Note that we didn't treat direct limits, so I pretty much only know about their definition in the category of topological spaces/sets.
In the notes, $X$ is a CW-complex, and the $X_i$ are their $i$-dimensional subcomplexes. If that makes things simpler, this can also be used as far as I care.
However, I don't want to use anything based on the fact that the $n$-th (relative) homology group of the cellular chain complex belonging to $X$ is isomorphic to $H_n(X)$, because the above will be used to prove this fact.

I'm suspecting that the proof needs more 'machinery' than we had been given. There are questions here on SO about when the homology functor commutes with the direct limit, but I didn't find helpful what I saw. I appreciate references, too, unless they refer to Hatcher's AT (not my style at all).
I'm somewhat familiar with and like the style Tammo tom Dieck's 'Algebraic Topology', for instance. But we didn't treat a lot of the concepts used in his book that would be needed to prove this, I think; and I don't have the time to work through all chapters preceding the pages where he uses the fact I want to prove here.

I'm wondering if one can prove the above in a very elementary way, i.e. not using much more than the definition of the topological direct limit (and I guess, using the direct limit of modules/groups would be necessary, too).
If less elementary concepts are needed, I would appreciate a proof as well, but maybe not straying too much from what is to be shown here.


For cellular subcomplexes we can use the fact that if there is a map $Y\to X$ and $Y$ is compact, then the image of $Y$ lies in finite union of $X$'s cells. (I tried to think out abstract condition for filtration $\{X_i\}$, but i failed.) The proof using this fact is short:

Let $\alpha\in C_n(X)$ be a cycle. Since $\alpha$ is a collection of singular simplexes which are compact, all these simplices lie in some $X_k$. So the element $[\alpha]\in H_n(X)$ comes from $H_n(X_k)$ (we may assume that $k>n$), thus the map $H_n(X_k)\to H_n(X)$ is a surjection.

Injectivity of this map can be shown in the same way.

In general the statement is false. Counterexample: let $X=S^1$ and $X_j=[0;2\pi-\frac1j]$. For this filtration we have $H_1(X_i)=0$ but $H_1(X)=\mathbb Z$.

  • $\begingroup$ What is a chain subcomplex? And what is 'the' map $\mathrm{colim} H_n(X_i) \rightarrow H_n(X)$ (I'm not really familiar with colimits of chains of modules)? $\endgroup$ – polynomial_donut May 6 '17 at 22:31
  • $\begingroup$ It would also be great if you could refer me to a definition for the colimit of a sequence of modules... I guess the colimit comes from: $X_i \hookrightarrow X_{i+1}$ is a topological embedding, so we have a sequence of injections $H_n(X_i) \overset {\iota_i} \hookrightarrow H_n(X_{i+1}) \hookrightarrow H_n(X_{i+1}) \hookrightarrow \dots$ and we define $\mathrm{colim}_i H_n(X_i)$ as $(\bigsqcup_i H_n(X_i))/ \sim$ with $\sim$ being generated by the requirement $X_j \ni x \sim \iota_j(x)$ for all $j$ and define the group operation accordingly (?). $\endgroup$ – polynomial_donut May 6 '17 at 22:50
  • $\begingroup$ @polynomial_donut colimit is just a synonym of a direct limit. the maps $H_n(X_i)\to H_n(X)$ may be not injective/surjective in general. $\endgroup$ – Andrey Ryabichev May 7 '17 at 8:47
  • $\begingroup$ I never mentioned that $H_n(X_i) \mapsto H_n(X)$ would be injective/surjective. I haven't really read anything about direct limits yet, either. But I assume it is what I am suggesting in my comment... ? $\endgroup$ – polynomial_donut May 7 '17 at 9:29
  • 1
    $\begingroup$ yes, your definition is right. $\endgroup$ – Andrey Ryabichev May 7 '17 at 9:40

Andrey Ryabichev gave me the important 'nudge' into the right direction (without which I might still be pondering about a too complicated approach to this proof, so thanks to him!), but I would like to 'flesh out' the proof he has given.

Surjectivity: Let $\zeta \in C_n(X)$ be an $n$-cycle. Then it is the finite formal sum of singular simplices. Those have compact image, so the finite union of their image lies in some $X_j$ for w.l.o.g. $j \ge n + 1$ (compact subsets of CW Complexes are contained in finite subcomplexes).
We get that $ C_n(X_j \overset \subset \longrightarrow X) $ restricts to a well-defined surjective homomorphism $$ Z_n(X_j) \longrightarrow Z_n(X)\,, $$ and considering that we have an injection of $\mathbb{Z}$-chains $ C_*(X_j) \overset {C_*(X_j \subset X)} \hookrightarrow C_*(X) $ is a morphism of chain modules, we see that this homomorphism induces a surjective homomorphism $$ C_n(X_j)/B_n(X_j)=H_n(X_j) \overset {H_n(X_j \subset X)} \longrightarrow H_n(X) = C_n(X)/B_n(X)\quad . $$

Injectivity: Assume that $H_n(X_j \subset X)([\zeta]) = [\zeta] =0\,$, implying that $\zeta$ represents a boundary in $H_n(X)$ , i.e. $\zeta = \partial_{n+1}(\beta)$ for, w.l.o.g., a singular $(n+1)$-simplex $\beta \in C_{n+1}$. By the same argument as above, we find some index $j_+ \ge j$ so that $\beta \in C_{n+1}(X_{j_+})$.

Consider the following diagram of abelian groups, where the horizontal morphisms are induced by the inclusions:

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} % C_{n+1}(X_j) & \ra{C_{n+1}(X_j \subset X_{j_+})} & C_{n+1}(X_{j_+}) & \ra{C_{n+1}(X_{j_+} \subset X)} & C_{n+1}(X) \\ \da{\partial_{n+1}} & & \da{\partial_{n+1}} & & \da{\partial_{n+1}}\\ C_{n}(X_j) & \ra{C_n(X_j \subset X_{j_+})} & C_{n}(X_{j_+}) & \ra{C_n(X_{j_+} \subset X)} & C_{n}(X) \\ % \end{array} $$

Since inclusions induce chain-maps under the functor $C_*: \mathrm{Top} \longrightarrow \mathrm{Ch}_\mathbb{Z}$, the above diagram commutes, and $C_n(X_j \subset X)$ is the composition of the morphisms in the lower row. But then, by commutativity, with $\chi:=C_n(X_j \subset X_{j_+})(\zeta)$, $$ C_n(X_{j_+} \subset X) (\chi) = C_n(X_j \subset X)(\zeta) = \partial_{n+1}(\beta) \circ C_{n+1}(X_{j_+} \subset X)= C_{n}(X_{j_+} \subset X) \circ \partial_{n+1}(\beta) \\ \implies \chi = \partial_{n+1}(\beta) $$ so if we consider the corresponding diagram chase inside singular homology (i.e., considering the quotient maps that define singular homology and are induced by the above maps by how singular homology is defined), this implies $H_n(X_{j+}) \ni [\chi] = 0$ , and using the given isomorphism $\varphi:=H_n(X_j \subset X_{j_+})$, this shows that $0 = \varphi^{-1}([\chi]) = \varphi^{-1}\left(\overline{C_n(X_j \subset X_{j_+})(\zeta)}\right)=\varphi^{-1} \varphi ([\zeta]) = [\zeta]$, where the last equation simply follows from the definition of the singular homology functor.


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