TITLE: Hatcher: Decomposition of space into X path components $\{X_\alpha\}$ gives isomorphism $H_n(X) \cong \bigoplus H_n(X_\alpha)$
QUESTION [6 upvotes]: 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?

REPLY [9 votes]: 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
$$H_n(X) = \ker(\partial_{n-1}^X)/\text{im}(\partial_n^X) \approx \ker \left(\bigoplus \partial_{n-1}^{X_\alpha} \right) / \text{im} \left(\bigoplus \partial_n^{X_\alpha} \right) \\ \approx \left(\bigoplus \ker (\partial_{n-1}^{X_\alpha}) \right) /  \left(\bigoplus \text{im}(\partial_n^{X_\alpha} )\right) \approx \bigoplus \ker (\partial_{n-1}^{X_\alpha}) / \text{im}(\partial_n^{X_\alpha} ) = \bigoplus H_n(X_\alpha).$$
Edited:
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
$$\ker(\partial_{n-1}^X)/\text{im}(\partial_n^X) = \phi_n(\ker(\bigoplus\partial_{n−1}^{X_\alpha})) / \phi_n(\text{im}(\bigoplus \partial_{n−1}^{X_\alpha})) \approx \ker \left(\bigoplus \partial_{n-1}^{X_\alpha} \right) / \text{im} \left(\bigoplus \partial_n^{X_\alpha} \right) $$