Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $A, B$ be two abelian categories, and $\tau : A \to B$ a left exact functor.

We define a category $C$ as follows:

objects: triples $(M, N, \varphi)$ where $M\in A, N\in B$ and $\varphi: N\to \tau M$ is a morphism in $B$.

morphisms: pairs $(f, g): (M, N, \varphi) \to (M^{\prime}, N^{\prime}, \varphi^{\prime})$ where $f:M\to M^{\prime}$ and $g:N\to N^{\prime}$ satisfying $(\tau f) \circ \varphi = \varphi ^{\prime} \circ g$.

Then, is the following statement true?

If so, then how can one prove it?

STATEMENT: If $A, B$ have enough injectives, then so does $C$.

For example, let $X$ be a scheme, $Y$ be a closed subscheme of $X$, and $U=X\setminus Y$. If $A$=(etale sheaves on $X$), $B$=(etale sheaves on $U$), then the cagegory $C$ is equivalent to the category of etale sheaves on $Y$. So, $C$ has enough injectives , of course. I wonder whether this kind of situation happens in the general setting above.

Please give me any advice.

share|improve this question
1  
You could break this up into two steps. The first, is whether for an abelian category with enough injectives, the arrow category has enough injectives. The second is to ask whether the category of abelian categories with enough injectives has strict pullbacks (I suspect yes, but I haven't thought about it for more than a few seconds). –  David Roberts Jun 18 '12 at 15:46
    
For the existence of cokernels, don't you also want to require that $\tau$ is right exact? –  Martin Brandenburg Jun 18 '12 at 18:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.