Theorem 5.13 in Rotman's Algebraic topology book from the chapter of homology exact sequences reads the following: Suppose, $X$ is a topological space with set of all path components, $\{X_\lambda:\lambda \in \Lambda\}$ and $A\subset X$. Then I am trying to prove that $H_n(X,A) \cong \oplus _{\lambda \in \Lambda} H_n(X_{\lambda}, A\cap X_{\lambda})$.
As the proof is left to the reader, I am writing the detailed proof that I have done and seek for verification of my proof. I am using set up in singular homology and following the notations given in Rotman.
$\mathbf{Lemma \ 1:}$ $S_n(X)\cong \oplus S_n(X_{\lambda})$ for all $n$. (Here $S_*$ denotes the singular complex of X. And by $S_*(A)$ we will mean the singular chain complex of A, regarded as a subcomplex of $S_*(X)$ that is, the generating $n$-simplexes will have its image inside $A$.)
$\mathbf{Proof:}$ For any $\gamma = \sum m_i\sigma_i \in S_n(X)$ we know that, each $\sigma_i$ has its image in some unique $X_{\lambda}$. So, clubbing the terms having its image inside a fixed $X_\lambda$ for each $\lambda$ together, we get a unique expression $\gamma= \sum \gamma_{\lambda}$ (uniqueness follows from the expression as a linear combination of it's basis elements).
So, the map $S_n(X) \rightarrow \oplus S_n(X_\lambda)$ sending $\gamma$ to $(\gamma_\lambda)$ is a well-defined group homomorphism having its inverse $(\gamma_\lambda)$ mapping to $\sum \gamma_\lambda$, which completes the proof.
$\mathbf{Reamrk:}$ We also see that this map sends $S_n(A)$ to $\oplus S_n(A \cap X_\lambda)$ isomorphically, since, if $\gamma$ is in $S_n(A)$ then image of $\sigma_i$'s is in A. Hence, image of basis elements in $\gamma_\lambda$'s lies in $X_\lambda \cap A$. Also the converse statement holds true.
Also, note that, this isomorphism is an isomorphism of Chain complexes $S_*(X) \cong \oplus S_*(X_\lambda)$ Also, induced map is an isomorphism of $S_*(A) \cong \oplus S_*(A \cap X_\lambda)$
$\mathbf{Lemma 2:}$ $H_n(\oplus S_*^\lambda)\cong \oplus_\lambda H_n(S_*^\lambda)$ , where $S_*^\lambda$ are chain complexes of abelian groups.
$\mathbf{Proof:}$ We can easily show, $Z_n(\oplus S_*^\lambda) = \oplus Z_n(S_*^\lambda)$ and $B_n(\oplus S_*^\lambda) = \oplus B_n(S_*^\lambda)$ for all n, Then, $H_n(\oplus S_*^\lambda) = Z_n(\oplus S_*^\lambda)/B_n(\oplus S_*^\lambda)=(\oplus Z_n(S_*^\lambda))/(\oplus B_n(S_*^\lambda))\cong\oplus(Z_n(S_*^\lambda)/B_n(S_*^\lambda))=\oplus H_n(S_*^\lambda)$.
Also, We note another result which says, if $\phi: G\rightarrow G'$ is an isomorphism of groups and $H,H'$ are normal subgroups of $G, G'$ respectively, such that $\phi(H)=H'$ Then $\phi$ induces an isomorphism of $G/H \rightarrow G'/H'$ sending $xH$ to $\phi(x)H'$.
( I guess the appropriate restatement of this result is true in the category of chain complexes of abelian groups : If we have an isomorphism of chain complexes of abelian groups $\phi:S_*\rightarrow S_*'$ if $A_*$ is a subcomplex of $S_*$ then, $A_* \cong im(\phi)$ and $S_*/A_* \cong S_*'/im(\phi)$ by the induced maps.)
Using these results, we observe from Lemma 1 and the remark following it,
$S_*(X)/S_*(A) \cong (\oplus S_*(X_\lambda))/(\oplus S_*(A \cap X_\lambda))\cong \oplus (S_*(X_\lambda)/S_*(A \cap X_\lambda))$
isomorphic as chain complexes.
$\therefore$ $H_n(X,A)= H_n(S_*(X)/S_*(A)) \cong H_n(\oplus (S_*(X_\lambda)/S_*(A \cap X_\lambda))) \\ \cong \oplus H_n(S_*(X_\lambda)/S_*(A \cap X_\lambda))=\oplus H_n(X_\lambda, A\cap X_\lambda)$.
Is this proof correct?
Is there any alternative way to prove this result?