Exact sequence $0\rightarrow H^i(A_{\bullet})\rightarrow\text{coker}f^{i-1}\rightarrow\text{im}f^i\rightarrow0$ - abstract nonsense proof? I'm a little stuck on an exercise about cohomology in an abelian category. Given a complex $A^{\bullet}$, where $f^i:A^i\rightarrow A^{i+1}$ (so $f^{i+1}f^i=0$), let $H^i(A^{\bullet}):=\text{coker}(\text{im}f^i\rightarrow\ker f^{i+1})$. I want to show there is an exact sequence $0\rightarrow H^i(A^{\bullet})\rightarrow\text{coker}f^{i-1}\rightarrow\text{im}f^i\rightarrow0$. I'm pretty sure I know where the maps come from but i'm not sure how to show exactness.
Warning: I'm new to homological algebra so I currently use some notation in order to keep track of things (and because I'm lazy when it comes to variable names). Notation for a morphism $f:A\rightarrow B$: the kernel of $f$ is $f_k:\ker f\rightarrow A$, and the cokernel $f^k:B\rightarrow\text{coker}f$. The image $\text{im f}\rightarrow B$ is thus $(f^k)_k$ but I simply denote it by $f_{im}$, and the map $A\rightarrow\text{im} f$ by $\tilde{f}$ (so $f=f_{im}\tilde{f})$.
In detail + my current progress: we have this commutative diagram. Where these maps came from: the ones with an $f$ in their labels are all apparent  (hopefully), and those with mono/epi arrow notation should also be mono/epi for clear or at least well-known reasons.
The map $g$ comes from the fact that $0=f^if^{i-1}=f^i(f^{i-1})_{im}\widetilde{f^{i-1}}$, so since $\widetilde{f^{i-1}}$ is epi, $f^i(f^{i-1})_{im}=0$ and thus $(f^{i-1})_{im}$ factors uniquely through $(f^i)_k$. Furthermore, $g$ is mono because $(f^i)_kg$ is (so if $g\alpha=0$ then $(f^i)_kg\alpha=0\Rightarrow\alpha=0$). The map $g^k$ is by definition the cokernel of $g$ and is epi for that reason.
Similarly, $0=f^if^{i-1}=(f^i)_{im}\widetilde{f^i}f^{i-1}$ and $(f^i)_{im}$ is mono, so $\widetilde{f^i}f^{i-1}=0$. Hence $\widetilde{f^i}$ factors uniquely through $(f^{i-1})^k$ - this is where $h$ comes from. And since $\widetilde{f^i}=h(f^{i-1})^k$ is epi, $h$ must be as well, which gives me the right side of the desired short exact sequence.
The real mystery lies with the morphism $H^i(A^{\bullet})\rightarrow\text{coker}f^{i-1}$. I constructed this morphism by noticing $0=(f^{i-1})^kf^{i-1}=(f^{i-1})^k(f^i)_kg\widetilde{f^{i-1}}$, so since $\widetilde{f^{i-1}}$ is epi, $(f^{i-1})^k(f^i)_kg=0$, so $(f^{i-1})^k(f^i)_k$ must factor through $g^k$, so we get $h'$.
At this point the duality/symmetry between the top left and bottom right side of the diagram is clear to me (indeed it looks as if $H^i(A^{\bullet})$ could also have been defined by $\ker(\text{coker}f^{i-1}\rightarrow\text{im}f^i)$ - is this right?) but I'm having trouble proving this symmetry exists. I can't even show $H^i(A^{\bullet})\rightarrow\text{coker}f^{i-1}$ is mono, let alone a kernel of $h$. One thing I can show is $hh'=0$: since $(f^i)_{im}hh'g^k=f^i(f^i)_k=0$, and since $g^k$ and $(f^i)_{im}$ are epi and mono respectively, we have $hh'=0$. I also rather suspect $h'g^k$ is the unique decomposition of $(f^{i-1})^k(f^i)_k$ through its image, but this is just an intuitive guess.
Now I saw this question but the answer appears to be appealing to the snake lemma. This is the second sequence in exercise 1.6.A on page 51 of Ravi Vakil's notes, and while the snake lemma is mentioned before this exercise, it is neither stated nor proven, so I'm wondering if there's a way to do without it. I can do an element chase, but I'm also wondering if there's an abstract nonsense way to finish the construction I've given.
 A: Note that the exact sequence is describing cohomology as a kernel instead of a cokernel. That is, you have a map
$f^j : A^j \longrightarrow A^{j+1}$
but since the image of $f^{j-1}$ is in the kernel of this map, you can consider
$$\bar f^j : A^j/\mathrm{Im}(f^{j-1}) = \mathrm{Coker}(f^{j-1}) \longrightarrow A^{j+1}$$
The induced map  $\bar f^j : A^j/\mathrm{Im}(f^{j-1}) = \mathrm{Coker}(f^{j-1}) \longrightarrow \mathrm{Im}(f^j)$ is surjective. Moreover, the kernel of $\bar f^j$ is $\mathrm{Ker}(f^j)/\mathrm{Im}(f^{j-1})= H^j(A)$, giving the usual short exact sequence
$$0 \longrightarrow H^j(A)  \longrightarrow   \mathrm{Coker}(f^{j-1}) 
  \longrightarrow  \mathrm{Im}(f^j)   \longrightarrow  0$$
of the form $0\to \mathrm{Ker}(\alpha) \to A \to \mathrm{Im}(\alpha)\to 0$ for a morphism $\alpha : A\to B$.
A: Here’s another proof, although it relies on a few results. Using the notation from the diagram in the question, we show two things:

*

*$h$ is the cokernel of $h’$

*$h’$ is mono

Then $h’$ is the kernel of $h$ and its own image since monos are the kernels of their cokernels in abelian categories, so we have the short exact sequence $0\rightarrow H^i(A^{\bullet})\rightarrow\text{coker}f^{i-1}\rightarrow\text{im}f^i\rightarrow0$.
To show the first point, suppose $x:\text{coker}f^{i-1}\rightarrow X$ such that $xh’=0$. Then $x(f^{i-1})^k(f^i)_k=xh’g^k=0$, and since $A^i\rightarrow\text{im}f^i$ is the coimage of $f^i$, it is the cokernel of $(f^i)_k$, so $x(f^{i-1})^k$ factors uniquely through $\widetilde{f^i}$, i.e. there is a unique morphism $y:\text{im}f^i\rightarrow X$ such that $x(f^{i-1})^k=y\widetilde{f^i}=yh(f^{i-1})^k$. Since $(f^{i-1})^k$ is epi, $x=yh$ with $y$ unique. Thus $h$ is the cokernel of $h'$.
Finally, to show $h'$ is mono, we can use this result, with $A=\text{im}f^{i-1}$, $B=\ker f^i$, and $C=A^i$.
