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Let $X$ be a topological space and $\{X_\alpha\}_{\alpha\in I}$ be the family of connected components of $X$.

Let $C:Top\rightarrow Ch(Ab)$ be the singular complex functor and $H_•:Top\rightarrow Ab$ be the singular homology functor.

I have proven that $C_n(X)\cong \oplus_{\alpha} C_n(X_\alpha)$, but I don't get why this implies that $H_n(X)\cong \oplus_{\alpha} H_n(X_\alpha)$. Could someone show me a concrete proof for this?

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  • $\begingroup$ I believe this is done in both Hatcher and Switzer's books, but the main idea is that these functors are nice enough to preserve products/coproducts (and yeah, your job here is to prove that!), and that X is the coproduct (disjoint union, for unpointed spaces) of its connected components, and these functors preserve that. $\endgroup$ Oct 1, 2016 at 1:33
  • $\begingroup$ To further comment, I think it's clear once you've found that fact out about the chain complex, because then the maps you build your homology out of are now direct sums of maps into the connect components, which should make it clear what to do. I'm not going to write an answer to this as it will take much longer to tex than it's worth (and its worthwhile for you to write this all down at least once in your life...) $\endgroup$ Oct 1, 2016 at 1:47
  • $\begingroup$ Maybe this is useful math.stackexchange.com/questions/1808392/… $\endgroup$
    – MathsIsFun
    Oct 1, 2016 at 6:48
  • $\begingroup$ Can you put it here how did you prove $C_n(X) \cong \oplus_{\alpha} C_n(X_{\alpha}?$ It will be really helpful. Thanks! $\endgroup$
    – Error 404
    Oct 9, 2018 at 14:29

2 Answers 2

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There is actually a short and simple answer since you have already proved that $C_n(X)\cong \oplus_a C_n(X_a)$.

Indeed, what you call the singular homology functor $H_\bullet :Top \to Ab$ is the composition of the singular complex functor $C: Top\to Ch(Ab)$ with the homology functor $H'_\bullet: Ch(Ab)\to Ab$ (for a presentation of this last functor see my answer to this question homology invariance). But a functor (here $H'_\bullet$) maps any iso (here $C_n(X)\cong \oplus_a C_n(X_a)$) to an iso, hence $H'_n(C_n(X))\cong H'_n(\oplus_a C_n(X_a))$. You need also to know the basic fact that $H'_n$ commutes with direct sum, meaning that $H'_n(\oplus_a C_n(X_a))\cong \oplus_a H'_n(C_n(X_a))$, so $H'_n(C_n(X))\cong \oplus_a H'_n( C_n(X_a))$, namely $H_n(X)\cong \oplus_a H_n(X_a)$.

In the following I prove the fact mentioned above, namely that the homology functor preserves coproducts (ie direct sums). For the sake of notations below I drop the ' and denote by $H_\bullet$ the homology functor from $Ch(Ab)$ to $Ab$.

Indeed, it suffices to prove that $H_n(\oplus A_a)$ has the universal property of $\oplus H_n(A_a)$ for any family $\lbrace A_a\rbrace$ of complexes in $Ch(Ab)$. To achieve this, it suffices to produce a universal cocone, ie a universal family of morphisms $\lbrace c_a:H_n(A_a)\to H_n(\oplus A_a)\rbrace$. Let us define $c_a$ by $H_n(c'_a)$ where $\lbrace c'_a:A_a\to\oplus A_a\rbrace$ is the universal cocone of $\oplus A_a$. Hence we have a cocone $\lbrace c_a\rbrace$, we need to prove it is a universal one. Let $G$ be an abelian group equipped with a cocone $\lbrace g_a:H_n(A_a)\to G\rbrace$. One needs to produce a morphism $\phi:H_n(\oplus A_a) \to G$, unique with the property that $\phi\circ c_a = g_a$. Let us define a complex $G_\bullet$ and a family of morphisms of complexes $\lbrace A_a\to G_\bullet \rbrace$ such that $H_n(G_\bullet)=G$ and $H_n(A_a\to G_\bullet)=g_a$ as follows. The complex $G_\bullet$ has only $G$ in its n-th spot and 0 elsewhere, and the morphism from $A_{a,n}$ to $G$ that maps $x$ to $g_a([x])$ gives us a morphism from $A_a$ to $G_\bullet$ with the required property. By the universal property of $\oplus A_a$ one gets a morphism $\phi'$ such that $\phi'\circ c'_a$ is equal to our morphism from $A_a$ to $G_\bullet$. Take $\phi = H_n(\phi')$, it exhibits our cocone $\lbrace c_a\rbrace$ as universal.

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  • $\begingroup$ The critical part is to show that the homology functor $H_n'$ commutes with direct sum. And this is exactly where I'm stuck at.. $\endgroup$
    – Rubertos
    Oct 1, 2016 at 11:31
  • $\begingroup$ ok, I will edit my answer this evening (when I will have free time) by writing a proof (at least a sketch) of this fact. $\endgroup$
    – A. Bordg
    Oct 1, 2016 at 11:34
  • $\begingroup$ thank you so much! I will try to figure out how until then.. $\endgroup$
    – Rubertos
    Oct 1, 2016 at 11:35
  • $\begingroup$ @Rubertos: done. $\endgroup$
    – A. Bordg
    Oct 1, 2016 at 13:14
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You use essentially the same argument as you did for $C_n(x) \cong \oplus_{\alpha} C_n (X_\alpha)$. The idea is that the boundary map $\partial_n(\sigma^n):= \sigma^n\vert_{\partial \Delta^n}$ is simply a restriction of the singular complex to its faces, which is also path connected (a simplex of dimension $n-1$.) Hence, the image of the singular simplex must be fully contained in a single path connected component. So, the boundary operator preserves this decomposition, meaning that $\ker \partial_n$ and $\textrm{Im} \, \partial_n$ also split as direct sums.

This is proposition 2.6 in Hatcher.

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  • $\begingroup$ I have no problem with showing that $C:Top\rightarrow Ch(Ab)$ preserves coproduct. My problem is to show that the homology functor $H_n':Ch(Ab)\rightarrow Ab$ preserves coproduct.. which you regard it as obvious $\endgroup$
    – Rubertos
    Oct 1, 2016 at 11:48

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