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 $im(f) \rightarrow ker(g)$.


  • $\begingroup$ A question that only makes sense with a specific book opened in a specific pag is not a good question. $\endgroup$ – Mariano Suárez-Álvarez Jan 18 '18 at 0:00
  • $\begingroup$ the book is available online - math.stanford.edu/~vakil/216blog/FOAGnov1817public.pdf I will edit it in my question $\endgroup$ – David Lui Jan 18 '18 at 0:11
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    $\begingroup$ 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. $\endgroup$ – 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).

  • $\begingroup$ Thank you. Is there some reason you avoided using ker(something) and coker(something), instead choosing to use letters such as K and L? Does this imply kernels are not unique? $\endgroup$ – David Lui Jan 19 '18 at 1:18
  • $\begingroup$ (Co)kernels are only unique up to isomorphism, and I also wanted to avoid confusion between the (co)kernel object and the (co)kernel morphism. $\endgroup$ – Eric Wofsey Jan 19 '18 at 2:08

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