Homology homomorphism of the inclusion map of a path component I'm quite new learning singular homology and from problem 9.9 from Greenberg & Harper's book of algebraic topology I ran into this problem:
Let $X$ to be a path component of $X'$ and let $\iota:X\hookrightarrow X'$ be the inclusion map. Then the induced homomorphism $H_{n}(\iota):H_{n}(X)\to H_{n}(X')$ is actually a monomorphism, for all $n\geq 0$.
The problem also asks to prove the existence of a homomorphism such that cancels $H_{n}(\iota)$ by the left but to be honest, where I'm stuck is proving the first part. Any advise will be appreciated.
 A: Let us first prove a lemma:
Lemma: Let $G,G',H$ be abelian groups and $\varphi : G \to H$ and $\varphi' : G' \to H$ be homomorphisms. If the map
$$
\varphi + \varphi' : G \oplus G' \to H, \qquad (g,g') \mapsto \varphi(g)+\varphi'(g'),
$$
is an isomorphism, then $\varphi$ and $\varphi'$ are injective maps.
Proof: Assume $\varphi$ is not injective. Then there exists some $g \in G \setminus \{0\}$ such that $\varphi(g)=0 \in H$. Then $(\varphi+\varphi')(g,0) = \varphi(g)+\varphi'(0) = 0+0=0$, but $(g,0)$ is non-zero in the abelian group $G\oplus G'$ and hence $\varphi+\varphi'$ is not injective. This contradicts the fact that $\varphi+\varphi'$ was an isomorphism.
Now we can prove that $H_n(\iota)$ is injective for all $n$: denote by $C$ the union of all path-components of $X$ that are not $X'$, i.e. $X = X' \cup C$. We have inclusions $\iota : X' \hookrightarrow X$ and $\iota_C : C \hookrightarrow X$ and $X' \cap C = \emptyset$. We know that $X',C$ are open in $X$, so we can use Mayer-Vietoris to obtain a long exact sequence
$$
\cdots\to H_n(\emptyset) \to H_n(X') \oplus H_n(C) \xrightarrow{\iota_*-(\iota_C)_*}H_n(X) \to H_{n-1}(\emptyset) \to \cdots
$$
Since $H_n(\emptyset) \cong 0$ for all $n$, we have an isomorphism $H_n(X') \oplus H_n(C) \xrightarrow{\iota_*-(\iota_C)_*}H_n(X)$. Now we can apply the above lemma with $\varphi = \iota_*$ and $\varphi' = -(\iota_C)_*$ to see that $\iota_* = H_n(\iota)$ is injective. I hope this was helpful!
