I am trying to compute the homology of the Klein bottle using the Eilenberg-Steenrod axioms and after a number of steps, I reached a part where I need to show that the map $f:S^1 \vee S^1\to S^1$, which leaves the left circle as is and collapses the right one onto the gluing point, induces the map $$f_* : H_1 (S^1\vee S^1)\to H_1(S^1),\qquad (a,b)\mapsto a$$ where we have identified $H_1(S^1\vee S^1)\cong \mathbb{Z}^2$ and $H_1(S^1)\cong \mathbb{Z}$. I have tried to compose $$S^1\overset{i}{\to} S^1\vee S^1\overset{f}{\to} S^1$$ where $i$ is the inclusion of $S^1$ into the left circle of $S^1\vee S^1$. Thus, since $H_1$ is a functor, we know that $f_*\circ i_*$ is the identity on $H_1(S^1)$. But now I'm stuck because in order to show that $f_*$ is the projection map, we need to show that $i_*$ maps $a\mapsto (a,0)$, which I am not sure how to proceed.
I'm aware that there are other ways of computing the Klein bottle that may not require this step. I am ultimately interested in proving this step though so if you could provide help as to how to prove this step, that'd be much appreciated.