Does $0\to \mathbb Z^2\to H_1(X,A)\to \mathbb Z^{k-1}\to 0$ split?

Question

Consider the following exact sequence: $0\to \mathbb Z^2\to H_1(X,A)\to \mathbb Z^{k-1}\to 0$.

How do I show that this splits?

By definition of splitting, I should have $H_1(X,A)= \mathbb Z^2 \oplus \mathbb Z^{k-1}=\mathbb Z^{k+1}$. But I don't understand how to show that this is the case here.

This came across while studying the relative homology groups $H_n(X,A)$ when $X$ is a torus and $A$ is a subcollection of $k$ points of $X$.

Answer 1

The most known result to show that a sequence splits is the splitting lemma.

Splitting Lemma. Let $$0\rightarrow A\overset{f}{\rightarrow} B\overset{g}{\rightarrow} C\rightarrow 0$$ be a short exact sequence in any abelian category. Then the following three statements are equivalent

1. There exists some left inverse $\tau:B\rightarrow A$ of $f$, that is $\tau\circ f=\text{id}_A$.
2. There exists some right inverse $\phi:C\rightarrow B$ of $g$, that is $g\circ\phi=\text{id}_C$.
3. The sequence splits, that is there is an isomorphism $\varphi:B\rightarrow A\oplus C$ such that the following diagram commutes. Hereby $i$ and $p$ are the canonical inclusion and projection respectively. $\require{AMScd}$ \begin{CD} 0 @>>> A @>f>> B @>g>> C @>>> 0\\ @. @V \text{id}_A VV @VV \varphi V @VV \text{id}_C V @.\\ 0 @>>> A @>i>> A\oplus C @>p>> C @>>> 0\\ \end{CD}

In addition to that the most basic general result you can use to show that your sequence splits probably is the following very useful Lemma.

Lemma 1. Every excat sequence of $\mathbb{Z}$-Modules $$0\rightarrow A\overset{f}{\rightarrow} B\overset{g}{\rightarrow} C\rightarrow0$$ with $C$ a free $\mathbb{Z}$-module splits.

To prove this statement we will need the following Lemma.

Lemma 2. Let $\varphi:A\rightarrow B$ be a homomorphism of $\mathbb{Z}$-modules. Further let $C$ be a free $\mathbb{Z}$-module. Then $\varphi$ is surjective if and only if for any homomorphism of $\mathbb{Z}$-modules $\psi:C\rightarrow B$ there is some $\mathbb{Z}$-module homomorphism $\Psi:C\rightarrow A$ such that $\varphi\circ\Psi = \psi$, that is the following diagram commutes. $$\require{AMScd}\begin{CD} A @<\exists \Psi << C \\ @V \varphi VV \swarrow_{\psi} \\ B \end{CD}$$

proof of Lemma 2. "$\Rightarrow$" As $C$ is free we can choose a basis $(c_i)$ of $C$. If $\varphi$ is surjective, we can pick $a_i\in\varphi^{-1}(\left\lbrace \psi(c_i)\right\rbrace)$. Now define $\Psi:C\rightarrow A$ by $\Psi(c_i):=a_i$. Then the above diagram obviously commutes.
"$\Leftarrow$" Take $C=\mathbb{Z}$. Since we can define $\psi(1):=b$ for all $b\in B$, $\varphi$ obviously has to be surjective.

proof of Lemma 1. By exactness the homomorphism $g:B\rightarrow C$ is surjective. Consider the identity map on $C$, $\text{id}_C:C\rightarrow C$. Then by Lemma 2 there exists a homomorphism $\Psi:C \rightarrow B$, such that the following diagram commutes. $$\require{AMScd}\begin{CD} B @<\exists \Psi << C \\ @V g VV \swarrow_{\text{id}_C} \\ C \end{CD}$$ In particular $g$ has a right inverse since $g\circ\Psi=\text{id}_C$. Thus by the splitting lemma the sequence splits.