Why is the homology group isomorphic to the homology groups of connected components

Question

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?

Answer 1

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.

Answer 2

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.