Proving there exists an isomorphism between the direct sum of the homology groups of path components and the homology group of the whole space 
Let $X$ be a space, let $\{X_\alpha\}_{\alpha\in A}$ be the set of path components of $X$, and let $\iota_{\alpha}:X_\alpha\to X$ be inclusion. Then for each $p\geq 0$, the maps $(\iota_\alpha)_*: H_p(X_\alpha)\to H_p(X)$ induce an isomorphism $$\bigoplus\limits_{\alpha\in A} H_p(X_\alpha)\to H_p(X)$$
Proof: Since the image of any singular simplex must lie entirely in one path component, it is clear that the chain maps $(\iota_\alpha)_\#: C_p(X_\alpha)\to C_p(X)$ already induce the isomorphism $$\bigoplus\limits_{\alpha\in A} C_p(X_\alpha)\to C_p(X)$$ The result for homology follows easily from this.

I don't understand parts of this proof.

*

*How does the fact that the images of simplices lie wholly in one path component imply that the maps $(\iota_\alpha)_\#: C_p(X_\alpha)\to C_p(X)$ induce an isomorphism $\bigoplus\limits_{\alpha\in A} C_p(X_\alpha)\to C_p(X)$?


*What exactly is this isomorphism? Does $(a,b,0,0,\dots)\to a+b$?
 A: 1) Consider $C_p(X_\alpha)$, which is the free abelian group with basis $\{\sigma:\Delta^p \to X_\alpha\}$ all singular simplices into $X_\alpha$. Any generator $\sigma\in C_p(X_\alpha)$ can be seen as a simplex in $X$ just by composing with $\iota_\alpha$. This means that the assignment $(\iota_\alpha)_\#:\sigma \mapsto \iota_\alpha \sigma$ maps a generator of $C_p(X_\alpha)$ to a generator in $C_p(X)$. Assembling together this assignment for every $\alpha\in A$ exhausts all of $C_p(X)$ and gives rise to an isomorphism $\phi:\bigoplus\limits_{\alpha\in A} C_p(X_\alpha)\to C_p(X)$ at the chain level:
Any element $x\in \bigoplus\limits_{\alpha\in A} C_p(X_\alpha)$ is a finite sum of the form $x=m_1\sigma_1 + ...+ m_k\sigma_k$ (each $\sigma_i:\Delta^p \to X_{\alpha_i}$) and maps by $\phi$ as $$x\mapsto \phi(x)=m_1(\iota_{\alpha_1}\sigma_1)+...+m_k(\iota_{\alpha_k}\sigma_k).$$
This is a well-defined homomorphism. In fact, clearly injective. Is it an isomorphism? Yes: Given any generator $\sigma:\Delta^p\to X$, as $\Delta^p$ is connected and $\sigma$ is continuous, $\sigma(\Delta^p)$ must be contained in some connected component, say $X_\beta$, so really $\sigma \in C_p(X_\beta)$, and $\sigma = \phi(\sigma)$, so $\phi$ is surjective. This is precisely the place where the claim "the image of a simplex lies in a path component" plays a role.
2) I think your sequence notation for direct sums is not very well adapted to the situation. In fact, if $X$ has uncountably many connected components, becomes confusing. It might be better in this context to think of formal sums, as I did above. Anyway, assuming $X$ has countably many connected components $X_1=X_{\alpha_1},...,X_k=X_{\alpha_k},...$ and assuming that the element $x=m_1\sigma_1 + ...+ m_k\sigma_k$ verifies that each $\sigma_i$ has image in $X_i$, then yes, precisely $$(m_1\sigma_1,...,m_k\sigma_k,0,...)\mapsto m_1\sigma_1 + ...+ m_k\sigma_k$$.
