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I am working on Exercise 5.21 from Rotman's algebraic topology book:

Assume that $X$ has five path components. If $CX$ is the cone on $X$, what is $H_1(CX,X)$?

Here is my solution:

We know that there is a long exact sequence $$\dots\to \tilde H_1(CX)\to H_1(CX,X)\to\tilde H_0(X)\to\tilde H_0(CX).$$ Since $CX$ is contractible, we know that $\tilde H_1(CX)=H_1(CX)=0$. This means that the image of the map $\tilde H_1(CX)\to H_1(CX,X)$ is $0$, and so the map $H_1(CX,X)\to\tilde H_0(X)$ has $\ker=0$.

We also know that $\tilde H_0(X)$ is the free abelian group with rank equal to one less than the number of path components of $X$. Thus $\tilde H_0(X)=\mathbb Z^4$ and $\tilde H_0(CX)=0$. Hence the map $H_1(CX,X)\to H_0(X)$ has image isomorphic to $\mathbb Z^4$ (since the map $\tilde H_0(X)\to\tilde H_0(CX)$ is everywhere $0$), from which we conclude that $H_1(CX,X)\cong\mathbb Z^4$.

I don't see any mistakes, but this is one of my first relative homology group computations, so I'd just like to check if the solution is actually correct.

EDIT: I just edited my solution (I messed up a bit by working with regular homology, instead of reduced homology groups; I think the solution is basically the same either way, but I think this one is better).

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This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct. If you have an exact sequence $$0 \to A \to B \to 0,$$ then $A \to B$ is an isomorphism. Take $A = H_1(CX,X)$ and $B = \tilde H_0(X)$.

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