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).


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 $B = \tilde H_0(X)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.