# 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 question that only makes sense with a specific book opened in a specific pag is not a good question. – Mariano Suárez-Álvarez Jan 18 '18 at 0:00
• the book is available online - math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf I will edit it in my question – David Lui Jan 18 '18 at 0:11
• Add the definitions of $f$ and $g$ and the other context. While the book being available online helps, it is much better for the question to be self-contained. – Derek Elkins left SE Jan 18 '18 at 0:19

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).