Let $(A^*, d^i)$ be a chain complex of finite dimensional vector spaces, i.e, $$0 \to A^0 \to A^1 \to \dots \to A^n \to 0.$$ Show the sequence $$0 \to H^i(A^*) \to A^i/Im(d^{i-1}) \to Im(d^i) \to 0$$ is exact. Here $H^i(A^*)$ denotes the $i$th cohomology group of $A^*$, that is we are required to show that $$0 \to \frac{Ker d^i}{Im (d^{i-1})} \to \frac{A^i}{Im (d^{i-1})} \to Im(d^i) \to 0 $$ is exact (for each $i$, but we show for a general $i$).
Can someone tell me why must the map $\frac{Ker d^i}{Im (d^{i-1})} \to \frac{A^i}{Im (d^{i-1})}$ be the inclusion map? We weren't told anything about the map $H^i(A^*) \to A^i/Im(d^{i-1})$.
I don't think your question has anything to do with topology. This is just a linear algebra question.
You have inclusion map $\ker(d^i)\to A^i$. Compose it with the projection $A^i\to A^i/Im(d^{i-1})$. By definition, the kernel of the resulting map is $\ker(d^i)\cap Im(d^{i-1})$.
But since $d^id^{i-1}=0$, you know that $Im(d^{i-1})\subset \ker(d^i)$, and so the intersection $\ker(d^i)\cap Im(d^{i-1})$ is just $Im(d^{i-1})$. Therefore, the kernel of the map $\ker(d^i)\to A^i\to A^i/Im(d^{i-1})$ is exactly $Im(d^{i-1})$. Then you can use the first isomorphism theorem to say that $\ker(d^i)/Im(d^{i-1})\to A^i/Im(d^{i-1})$ is isomorphism on its image, and therefore injective.