What is the monomorphism that forms the homology group? On page 50 of Vakil's algebraic geometry, it mentions there is a monomorphism $im(f) \rightarrow ker(g)$, where $f : A \rightarrow B$, $g : B \rightarrow C$ is a complex in an abelian category - that is, $g \circ f = 0$. What is this monomorphism? I can't even think of a good candidate for the function $im(f) \rightarrow ker(g)$.
http://math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf
 A: Intuitively, if our category was modules over some ring, then we have homomorphisms $f:A\to B$ and $g:B\to C$ such that $g\circ f=0$.  This means that $g(f(a))=0$ for all $a\in A$, so the image of the map $f$ is a subset of the kernel of the map $g$.  The natural map $\mathrm{im}(f)\to \ker(g)$ is then just the inclusion map.
In a general abelian category, you just have to find a way to describe this using the universal properties of images and kernels.  Vakil defines $\mathrm{im}(f)=\ker(\mathrm{coker}(f))$, so we'll have to use the universal properties of kernels and cokernels.  Let $p:B\to K$ be a cokernel of $f$.  Since $g\circ f=0$, there is a morphism $g':K\to C$ such that $g'p=g$.
Now let $i:L\to B$ be a kernel of $p$ and let $j:M\to B$ be a kernel of $g$.  Since $gi=g'pi=g'0=0$, there is a morphism $i':L\to M$ such that $ji'=i$.  This turns out to be exactly the map you want (note that $L=\ker(\mathrm{coker}(f))=\mathrm{im}(f)$ and $M=\ker(g)$, and you can check that in a category of modules over a ring, this is just the map described in the first paragraph).
