Problem: Let $0 \to A^0\stackrel{d^0}{\to} A^1 \stackrel{d^1}{\to} \dots \stackrel{d^{n-1}}{\to} A^n \to 0 $ be a chain complex and assume each $A^i$ is finite-dimensional. Prove that the sequence $$0 \to H^i(A^*) \to \frac{A^i}{Im(d^{i-1})} \stackrel{d^i}{\to} Im(d^{i}) \to 0 $$ is also exact. where $H^i(A^*) = \frac{Ker(d: \Omega^{p}(A^*) \to \Omega^{p+1}(A^*))}{Im(d: \Omega^{p-1}(A^*) \to \Omega^{p}(A^*))}$ where $A^* = (A^i, d^i)$ is precisely $$0 \to A^0\stackrel{d^0}{\to} A^1 \stackrel{d^1}{\to} \dots \stackrel{d^{n-1}}{\to} A^n \to 0 $$
The key says,
Q1: Why is the induced quotient map $\frac{Ker(d^i)}{Im(d^{i-1})} \to \frac{A^i}{Im(d^{i-1})}$ the inclusion map? No where does it say this really is the inclusion map.
Q2: Looks like $d^i$ is just a regular homomoprhism, it is not surjective, so why do they say $x = d^i(y)$?
Q3: What does "$d^{i}([x]) = d^i(x)$" even mean? The notations are confusing. How can a set $[x]$ equal an element $x$? Not to mention, I don't feel like they proved the subjectivity of $d^i$ at all…they just stated the definition. I can understand the exactness in the middle once I understand the notations.(answered) thanks
1) It's not necessarily true that $\frac{\ker(d^i)}{\text{Im}(d^{i-1})} \supseteq \frac{A^i}{\text{Im}(d^{i-1})}$, and we don't need this. The unlabeled map $\frac{\ker(d^i)}{\text{Im}(d^{i-1})} \to \frac{A^i}{\text{Im}(d^{i-1})}$ is supposed to be an inclusion, so it suffices to check that $\ker(d^i) \subseteq A^i$, since this inclusion will descend to the quotient.
2) $x$ is an element of $A^i$, while $[x]$ is an element of $A^i/\text{Im}(d^i)$, i.e., a coset $x + \text{Im}(d^i)$. So in order to define $d^i : A^i/\text{Im}(d^i) \to \text{Im}(d^i)$ by $d^i([x]) = d^i(x)$, we need to ensure that this definition is independent of the representative we choose for the coset $x + \text{Im}(d^i)$.
3) Technically, the $d^i$ on the left is not the same map as the $d^i$ on the right. The one on the left is a map $d^i : A^i/\text{Im}(d^i) \to \text{Im}(d^i)$ that we want to define, while the one on the right is $d^i : A^i \to A^{i+1}$. This is sloppy notation, but understandable considering how many maps we have lying around. Maybe we should call the map on the quotient $\overline{d^i} : A^i/\text{Im}(d^i) \to \text{Im}(d^i)$ and say that we want to define $\overline{d^i}$ by $\overline{d^i}([x]) = d^i(x)$.
Here is how their proof shows $\overline{d^i}$ is well-defined. Take a coset $[x] = x + \text{Im}(d^{i-1})$ and suppose we pick a different representative $y$ for this coset. Then $x + \text{Im}(d^{i-1}) = y + \text{Im}(d^{i-1})$, so $x - y + \text{Im}(d^{i-1}) =\text{Im}(d^{i-1})$, hence $x = y + d^{i-1}(z)$ for some $z \in A^{i-1}$. They show that $d^i(x) = {d^i}(y + d^{i-1}(z))$ and ${d^i}(y)$ are the same, hence the value we get is independent of the choice of representative for a coset. Thus the map $\overline{d^i}$ defined by $\overline{d^i}(x + \text{Im}(d^{i-1}) = d^i(x)$ is well-defined.