# Does the adjugant of additive functors between abelian categories preserve the abelian structure of the hom-set?

I think the following is a counter-example. I noticed it when trying to prove that the sheafification functor induces isomorphism on the stalks (Vakil 2.4M).

As in Vakil (2.6.3) the stalk functor is left adjoint to the skyscraper. So, let $\mathcal F$ be a presheaf of abelian groups and consider the adjunction $$\eta:\def\Hom{\text{Hom}\,}\Hom(\mathcal F^{sh}_p,\mathcal F_p) \xrightarrow{\sim} \Hom(\mathcal F^{sh},i_p(\mathcal F_p)),$$where $i_p$ denotes the skyscraper functor.

Assuming what I wanted to prove, namely that the sheafification functor induces isomorphims on the stalks, we get an isomorphism on the left-hand side being mapped to something which is certainly not an isomorphism for general $\mathcal F$ (unless it is also a skyscraper sheaf at $p$). What I mean is any map $\phi \in \Hom(\mathcal F^{sh},i_p(\mathcal F_p)),$ has a non-zero kernel (it includes all sections of $\mathcal F$ over open sets away from $p$).

1) How can my $\eta$ still send an isomorphism to something that has a kernel?

2) How can one use the adjointness of stalks and skyscraper to get the induced isomorphisms?

## 2 Answers

I will denote by $F'$ the sheaf associated to $F$ and by $i_p A$ the skyscraper sheaf of $A$ (an arbitrary abelian group) at $p$. From

$\hom(F'_p,A) \cong \hom(F',i_p A) \cong \hom(F,i_p A) \cong \hom(F_p,A)$

we get $F'_p \cong F_p$ (Yoneda).

Concerning the question in the title (which is a completely different one?!): If $C,D$ are linear categories (they don't have to abelian) and $F : C \to D$, $G : D \to C$ are linear functors such that $F$ is left adjoint to $G$ as a usual functor, then actually $F$ is left adjoint to $G$ as a linear functor, i.e. we have not just bijections $\hom(Fx,y) \cong \hom(x,Gy)$, but rather isomorphisms of abelian groups. The reason is that the map is given by $f \mapsto G(f) \circ \eta_x$, where $\eta_x : x \to F(G(x))$ is the unit morphism. This map is clearly additive.

• Thanks! 2 comments: 1) did you get the last iso in you answer to my second question from adjunction of stalks and skyscraper as functors from (and to) presheaves? I didn't realize they existed and were also adjoint in the larger category. Is this a general fact when one cat is a full subcat of another? 2) As to the question in the title: I appreciate your explanation/proof, but I'm still confused because then the iso $Hom(\mathcal F^{sh}_p,\mathcal F_p) \ni f \mapsto \text{sth that is not an iso}$, as anything in $Hom(\mathcal F^{sh},i_p(\mathcal F_p))$ has a kernel (everything outside $p$) – Rodrigo Oct 26 '13 at 14:05
• 1) Not a general fact, but here it's obviously true. – Martin Brandenburg Oct 26 '13 at 20:31
• Right, thank you! I also just understood the answer to my question 1), which I reformulated above. The fact that $\eta$ must map the $\text{id}$ morphism $\in \def\Hom{\text{Hom}\,}\Hom(\mathcal F^{sh}_p,\mathcal F_p)$, which is also the identity $1_{\Hom(\mathcal F^{sh}_p,\mathcal F_p)}$ as an element of the abelian hom-group, to $1_{\Hom(\mathcal F^{sh},i_p(\mathcal F_p))}$ does not mean that this image element will also be an iso (but it's still the identity in the latter group). – Rodrigo Oct 26 '13 at 21:14

Reading the question and comments on Martin's answer, I think the following is bothering you:

If one has an isomorphism induced by some natural adjunction of functors, how can it be that a morphism that is an isomorphism is identified with a morphism that has a non-trivial kernel?

Why don't you consider a simpler example of such an adjunction isomorphism, e.g. $$Hom(\mathbb Z, A) = Hom(\mathbb Z/2\mathbb Z,A),$$ whenever $A$ is an abelian group that is $2$-torsion, i.e. killed by mult. by $2$.

Firstly, you surely will have no trouble verifying this adjunction isomorphism.

Secondly, if you take $A = \mathbb Z/2$, then the identity isomorphism $\mathbb Z/2 \to \mathbb Z/2$ in the right-hand side is identified with the natural surjection $\mathbb Z \to \mathbb Z/2$ on the left-hand side, which is certainly not an isomorphism.

I'm not sure what more there is to say. You have a mistaken expectation about how things should go, but hopefully looking at this example, you can figure out why your expectation is mistaken, and revise your intuition accordingly.

(Actually, one thing that is related and which might help is to consider the exactness properties of the various functors involved, and some related things, but probably it is best to leave a bare-bones example first.)

• Yup. You had a good example! – Rodrigo Oct 26 '13 at 21:24
• Matt, I really enjoy reading your posts. – Martin Brandenburg Oct 26 '13 at 23:05
• @MartinBrandenburg: Dear Martin, Thank you very much. Best wishes, – Matt E Oct 27 '13 at 1:03
• What an awesome example! – user123454321 Feb 10 '14 at 13:37