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Consider the direct image functor $f_*: Sh(X) \rightarrow Sh(Y)$, let $X$ and $Y$ be topological spaces, let $f: X \rightarrow Y$ be a continuous map, let $G \in Sh(X)$ be a sheaf. I was reading this course on sheaves:

http://bit.ly/14IBBTZ.

On page 1 they define this subpresheaf $f_!G \subset f_*G$ then on page 2 in the first line they remark that:

"As $f_!$ is a subfunctor of $f_*$ it is left exact".

I wanted to ask, are subfunctors of left exact functors also left exact?

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    $\begingroup$ Yes, obviously. $\endgroup$
    – Steven Landsburg
    Commented Apr 1, 2013 at 16:26
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    $\begingroup$ It is obvious that a subfunctor of a left exact functor sends short exact sequences to sequences that are exact on the left (i.e., it sends injections to injections), but it is neither obvious nor true that it preserves exactness in the middle. $\endgroup$
    – Eric Wofsey
    Commented Apr 1, 2013 at 17:38
  • $\begingroup$ Eric: Aagh. You're right. $\endgroup$
    – Steven Landsburg
    Commented Apr 1, 2013 at 17:51
  • $\begingroup$ Thanks, does that mean that $f_!$ might not be left exact then? $\endgroup$
    – Louis A
    Commented Apr 2, 2013 at 5:20
  • $\begingroup$ It turns out that $f_!$ is still left exact. This follows from the fact that $f_*$ is left exact and for any subsheaf $\mathcal{F}\subset\mathcal{G}$, $f_!\mathcal{G}\cap f_*\mathcal{F}=f_!\mathcal{F}$ as subsheaves of $f_*\mathcal{G}$. $\endgroup$
    – Eric Wofsey
    Commented Apr 2, 2013 at 18:53

2 Answers 2

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Here's a counterexample with additive functors on abelian categories. If $A$ is an abelian group, let $F(A)$ denote the subgroup of elements that are divisible by $2$. It is easy to see that $F:Ab\to Ab$ is an additive functor, and $F$ is a subfunctor of the identity. But $F$ is not left exact because it does not preserve kernels. For instance, $F$ sends the exact sequence $$0\to \mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}$$ to $$0\to2\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z},$$ which is not exact.

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Not necessarily. Consider these two functors Set → Set: F is the constant functor with value the empty set, and G is the identity functor. Then F is a subfunctor of G, G is obviously left exact, but F is not: it does not send the terminal object to the terminal object.

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    $\begingroup$ Everyone can stop down-voting my answer: I've corrected the spelling of 'necessarily'. :P (Seriously now, I agree that Eric's example is much nicer and closer to the context Samuel Mf is interested in.) $\endgroup$
    – Omar Antolín-Camarena
    Commented Apr 2, 2013 at 15:30

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