This is not a duplicate. Other answers to this question used language of category theory, which I am not familiar to (I don't know what a functor is etc...)

Proposition 2.6 in Hatcher's book, p109, says:

Corresponding to the decomposition of a space $X$ into its pathcomponents $X_\alpha$, there is an isomorphism of $H_n(X)$ with the direct sum $\bigoplus_\alpha H_n(X_\alpha)$.

Proof: Since a singular simplex always has path-connected image, $C_n(X)$ splits as the direct sum of its subgroups $C_n(X_\alpha)$. The boundary maps $\partial_n$ preserve this direct sum decomposition, taking $C_n(X_\alpha)$ to $C_{n-1}(X_{\alpha})$, so $\operatorname{Ker} \partial_n$ and $\operatorname{Im} \partial_{n+1}$ split similarly as direct sums, hence the homology groups also split, $H_n(X) \cong \bigoplus H_n(X_\alpha)$. $\quad \square$

I managed to show that $C_n(X) \cong \bigoplus C_n(X_\alpha)$ but I'm stuck from there.

What does "The boundary maps $\partial_n$ preserve this direct sum decomposition, taking $C_n(X_\alpha)$ to $C_{n-1}(X_{\alpha})$" even mean? Do I have to prove that

$$\partial_n(C_n(X)) \cong \bigoplus_\alpha \partial_n(C_n(X_\alpha))$$?

If so, how do I show this?

Also, how does "$\operatorname{Ker} \partial_n$ and $\operatorname{Im} \partial_{n+1}$ split similarly as direct sums" then follow and why does this even imply that the homology groups split as direct sums?

  • $\begingroup$ The boundary maps preserve the direct sums means exactly what you wrote and even more - that $\partial_n C_n(X_\alpha)$ is in $C_{n-1}(X_\alpha)$ (not in any of the other $C_{n-1}(X_\beta)$'s. This gives the splitting of the kernel and images almost directly. $\endgroup$ – Michael Burr Jan 30 at 15:39
  • $\begingroup$ And how can I prove this? If you would just hint me the right isomorphism, this would already give me a good start! $\endgroup$ – user745578 Jan 30 at 15:41
  • $\begingroup$ It's really just from the definition. If you're using singular singular homology, you can see that since a simplex is path connected, its image is also path connected (under a continuous map). Now, apply this to the simplices that appear in the definition of the boundary map. $\endgroup$ – Michael Burr Jan 30 at 15:46
  • $\begingroup$ Sorry, I'm not really sure how that gives me an isomorphism. My definition is indeed $\partial_n(\sigma) = \sum_i (-1)^i \sigma\vert{[v_0, \dots, \hat{v_i}, \dots, v_n]}$ where $\sigma$ is an $n$-simplex. $\endgroup$ – user745578 Jan 30 at 15:49

If you have any subspace $X' \subset X$, then $C_n(X')$ canonically embeds as a subgroup of $C_n(X)$:

Let $j : X' \to X$ denote inclusion, then we define $C_n(j) : C_n(X') \to C_n(X)$ by $C_n(j)(\sigma) = j \circ \sigma$ on the generators $\sigma : \Delta^n \to X'$. This is obviuosly an embedding of free abelian groups. Consider the boundaries $\partial_n^{X'} : C_{n+1}(X') \to C_n(X')$ and $\partial_n^{X} : C_{n+1}(X) \to C_n(X)$. Clearly we have $C_n(j) \circ \partial_n^{X'} = \partial_n^{X} \circ C_{n+1}(j)$.

The isomorphism $\phi_n : \bigoplus C_n(X_\alpha) \to C_n(X)$ has therefore the property $$\phi_n \circ \bigoplus \partial_n^{X_\alpha} = \partial_n^{X} \circ \phi_{n+1} .$$ This is the meaning of "The boundary maps preserve this direct sum decomposition".

Thus $$H_n(X) = \ker(\partial_{n-1}^X)/\text{im}(\partial_n^X) \approx \ker \left(\bigoplus \partial_{n-1}^{X_\alpha} \right) / \text{im} \left(\bigoplus \partial_n^{X_\alpha} \right) \\ \approx \left(\bigoplus \ker (\partial_{n-1}^{X_\alpha}) \right) / \left(\bigoplus \text{im}(\partial_n^{X_\alpha} )\right) \approx \bigoplus \ker (\partial_{n-1}^{X_\alpha}) / \text{im}(\partial_n^{X_\alpha} ) = \bigoplus H_n(X_\alpha).$$


Concerning the first isomorphism in the above chain:

Let $a \in C_n(X)$. Then $a \in \ker(\partial^X_{n−1})$ iff $\partial^X_{n−1}(a)=0$ iff $(\phi_{n−1} \circ \bigoplus \partial_{n−1}^{X_\alpha} \circ \phi_n^{-1})(a)=0$ iff $\bigoplus \partial_{n−1}^{X_\alpha} (\phi_n^{-1}(a))=0$ iff $\phi_n^{-1}(a) \in \ker(\bigoplus \partial_{n−1}^{X_\alpha})$ iff $a \in \phi_n(\ker(\partial_{n−1}^{X_\alpha}))$, i.e. we have $\ker(\partial^X_{n−1})=\phi_n(\ker(\bigoplus \partial_{n−1}^{X_\alpha}))$. Similarly $\text{im}(\partial^X_{n−1})=\phi_n(\text{im}(\bigoplus \partial_{n−1}^{X_\alpha}))$. Hence $$\ker(\partial_{n-1}^X)/\text{im}(\partial_n^X) = \phi_n(\ker(\bigoplus\partial_{n−1}^{X_\alpha})) / \phi_n(\text{im}(\bigoplus \partial_{n−1}^{X_\alpha})) \approx \ker \left(\bigoplus \partial_{n-1}^{X_\alpha} \right) / \text{im} \left(\bigoplus \partial_n^{X_\alpha} \right) $$

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  • $\begingroup$ Thank you very much! Can you further explain how this implies that $H_n(X) \cong \bigoplus_\alpha H_n(X_\alpha)$? $\endgroup$ – user745578 Jan 30 at 17:36
  • $\begingroup$ See my edit ... $\endgroup$ – Paul Frost Jan 30 at 17:46
  • $\begingroup$ Thank you very much! I'll try to make sure I get it! $\endgroup$ – user745578 Jan 30 at 17:56
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    $\begingroup$ @user745578 This is just a special case of assigning to a map $f : X \to Y$ induced homomorphisms $C_n(f) = f_\# :C_n(X) \to C_n(Y)$, i.e. a chain map $C_*(f) : C_*(X) \to C_*(Y)$. See Hatcher p. 110/111. $\endgroup$ – Paul Frost Feb 10 at 23:07
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    $\begingroup$ I added my comment to my answer. $\endgroup$ – Paul Frost Apr 18 at 17:23

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