Setting: Let $\mathscr{C}\subset \mathscr{A}$ be a full additive subcategory of an abelian category. Let $C,C'$ be objects of $\mathscr{C}$ and let $f\in \operatorname{Hom}_\mathscr{C}(C,C')=\operatorname{Hom}_{\mathscr{A}}(C,C')$.


Suppose that $C''\in \mathscr{C}$ and that $i\in\operatorname{Hom}(C'',C)$ is a kernel of $f$ in the category $\mathscr{C}$. Is $i$ also a kernel of $f$ in $\mathscr{A}$?

I suppose the answer is yes (see context, below), but I would like to understand why.

I am taking all definitions from Weibel, Introduction to Homological Algebra. He defines a kernel $i$ of a map $f$ in an additive category as a map that is universal with respect to $f\circ i = 0$.

Context: Weibel states on p. 25 that a full subcategory of an abelian category is abelian (and the inclusion is exact) iff it is additive and closed under kernels and cokernels. At first I thought this was completely obvious, as an abelian subcategory must contain kernels and cokernels, and then their appropriate properties will be inherited from the ambient category. However, then I became concerned:

As there are more objects in $\mathscr{A}$ than $\mathscr{C}$, there are (potentially) many more maps that must factor through $i$ for it to be a kernel in $\mathscr{A}$ than just those that factor through it because it is a kernel in $\mathscr{C}$. Thus being a kernel in $\mathscr{A}$ is a more stringent requirement than being a kernel in $\mathscr{C}$.

When I tried to write down a direct proof from first principles, the fact that $\mathscr{A}$ is abelian lets me suppose there is an $A\in\mathscr{A}$ with $j:A\rightarrow C$ a kernel in the ambient category, and then there must exist $\pi\in\operatorname{Hom}_{\mathscr{A}}(C'',A)$ with $i=j\circ\pi$, but I can't see anything that lets me construct a map inverse to $\pi$.

I have similar concerns about cokernels, epis and monics. In all cases, the requirements to be one of these seem more stringent in $\mathscr{A}$ than in $\mathscr{C}$, so I don't see that the properties coincide in $\mathscr{A}$ and $\mathscr{C}$, in which case I am confused why to expect Weibel's claim to be true.

On the other hand, I assume it is true (because it's in the book), which inclines me toward the belief that the answer to my question is yes, because this seems like the most straightforward route to the truth of Weibel's claim.

So, what's the real situation? Thanks in advance for your time and thought.


No, this is not true. I will give a counterexample for cokernels; taking the opposite categories then gives an example for kernels. Let $\mathscr{A}=Ab$ and let $\mathscr{C}$ be the full additive subcategory consisting of torsion-free abelian groups. Consider the map $f:\mathbb{Z}\to\mathbb{Z}$ given by multiplication by $n$, for $n>1$. It is not hard to see that $\mathbb{Z}\to 0$ is a cokernel of $f$ in $\mathscr{C}$. However, it is not a cokernel of $f$ in $\mathscr{A}$, because the quotient $\mathbb{Z}\to\mathbb{Z}/n$ does not factor through it. This is also a counterexample for epics: the map $f$ is epic in $\mathscr{C}$, but not in $\mathscr{A}$.

The assertion in Weibel is actually weaker than you are claiming: he says that $\mathscr{C}$ is abelian and an exact subcategory of $\mathscr{A}$ iff it is additive and closed under kernels and cokernels. Here "exact subcategory" means exactly that kernels and cokernels in $\mathscr{C}$ are the same as kernels and cokernels in $\mathscr{A}$.

In fact, your version of Weibel's assertion is false in general. For instance, let $\mathscr{C}$ be any small abelian category and let $\mathscr{A}$ be the category of all additive functors $\mathscr{C}^{op}\to Ab$. Then $\mathscr{A}$ is abelian (kernels and cokernels are computed pointwise), and there is a functor $H:\mathscr{C}\to\mathscr{A}$ which sends $C$ to the functor $\operatorname{Hom}(-,C)$. By Yoneda's lemma, $H$ is fully faithful, so we may consider $\mathscr{C}$ as a full subcategory of $\mathscr{A}$. However, cokernels in $\mathscr{C}$ are usually not the same as cokernels in $\mathscr{A}$: cokernels in $\mathscr{A}$ are computed pointwise, and the functor $\operatorname{Hom}(B,-)$ does not preserve cokernels unless $B$ is projective, so $H$ preserves cokernels iff every object of $\mathscr{C}$ is projective.

  • $\begingroup$ +1 Thanks. ($\mathscr{A}$ is the category of abelian groups I take it.) So then Weibel's claim is true for a more subtle reason I guess... $\endgroup$ – Ben Blum-Smith Sep 18 '15 at 20:20
  • $\begingroup$ No, you're misreading Weibel's claim, and in fact your version of his claim is false in general. See the update to my answer above. $\endgroup$ – Eric Wofsey Sep 18 '15 at 21:47
  • $\begingroup$ Oh, this is very helpful. I had assumed that $\mathscr{C}$ abelian would by itself imply closure under kernels and cokernels and this would in turn imply both that $\mathscr{C}$ was abelian and also that the inclusion was exact. I didn't realize the importance of the assumption that the inclusion was exact as a hypothesis in the $\Rightarrow$ direction. $\endgroup$ – Ben Blum-Smith Sep 21 '15 at 2:43

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