Why is the inclusion an isomorphism?

Question

Consider $X$ a path-connected space, $A\subset X$ a non-empty subset.

My textbook makes the following claim without any explanation, and I wondered if you could help: it says that the inclusion $H_0 (A)\to H_0 (X)$ is onto.

Why is that true? It is easy to see that $H_0(A)\cong H_0(X)$ since they are both $\mathbb Z$, but why does the generator of $H_0(A)$ map to the generator of $H_0(X)$?

thanks

Answer 1

An element of $H_0(X)$ is basically just a point. Take any $a\in A\,,$ then $a\in X$ generates $H_0(X)$ since any other point $x\in X$ can be connected to $a$ since $X$ is path connected, hence $x-a$ is a boundary, ie. $x-a\equiv 0\,,$ hence $x\equiv a\,.$