Why exactly is this injective? Algebraic Topology. 
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 $H^i(A^*) \to A^i/Im(d^{i-1})$. 
 A: Showing that a potentially short exact sequence is exact without knowing what the maps are is a fruitless endeavor. If the writer did not explicitly give the maps, its likely they just meant the most obvious ones (in this case where the first map is the map induced from the inclusion and the second map is induced from $d^i$).
A: 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.
