I can't seem to figure out why $H_0(X) \cong \tilde{H}_0(X) \oplus \mathbb{Z}$. In my notes it says:

''...Since $\varepsilon \partial_1 = 0$, $\varepsilon$ vanishes on $im \partial_1$ and hence induces a map $H_0(X) \rightarrow \mathbb{Z}$ with kernel $\tilde{H}_0(X)$...''

How is this map induced? Why is $\mathbb{Z}$ relevant?. Many thanks for your help.


1 Answer 1


The map $H_0(X)\to\mathbb{Z}$ is induced by the map on chains $C_0(X)\to\mathbb{Z}$ sending each singular simplex to $1$. Since $\mathbb{Z}$ is a projective $\mathbb{Z}$-module (even free), you can always pick a spliting $\mathbb{Z}\to H_0(X)$. Therefore, $H_0(X)\simeq \mathbb{Z}\oplus\tilde{H}_0(X)$.


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.