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This might be a rather stupid question, however, I cannot make sense of what is going wrong.

Let $f:X\to Y$ be a morphism of ringed spaces, and $\mathcal{F,G,H}$ sheaves of abelian groups on $X$. Show that the direct image functor $f_*$ is left exact.

Suppose $0\to\mathcal F\to\mathcal G\to\mathcal H\to 0$ is a short exact sequence of sheaves. In particular, for every $U\subset X$, the sequence $0\to\mathcal F(U)\to\mathcal G(U)\to\mathcal H(U)\to 0$ is short exact.

Consider the sequence $0\to f_* \mathcal F\to f_* \mathcal G\to f_* \mathcal H\to 0$. This is left exact if the sequence $0\to f_* \mathcal F (V)\to f_* \mathcal G(V)\to f_* \mathcal H(V)\to 0$ is left exact for every $V\subset Y$ open. This latter sequence can also be written by definition as $0\to \mathcal F (f^{-1}(V))\to \mathcal G(f^{-1}(V))\to \mathcal H(f^{-1}(V))\to 0$. But since $f^{-1}(V)$ is open in $X$, then the previous sequence is already short exact.

What is wrong in this argument? I think I am missing something big.

George
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    It is not true that $0 \to F(U) \to G(U) \to H(U) \to 0$ is exact for every open $U \subset X$. In general, the $U$-section functor $F \mapsto F(U)$ is also only left-exact. – red_trumpet Jun 11 '19 at 06:24
  • I accidentally took that as the definition, thank you for pointing out my mistake. – George Jun 11 '19 at 12:44

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You will have exactness at the stalks, but in general you do not have exactness at sections over $U$.

Hagen von Eitzen
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  • I see how surjectivity at the stalks does not imply surjectivity on sections. However, I don't see why it is necessary to pass to stalks at all. – George Jun 11 '19 at 00:18
  • @George What is your definition of exactness, or more specifically, surjectivity? Recall that the image sheaf of a morphism $f$ is defined as the sheafification of the presheaf $U \mapsto \operatorname{img}(f|_U)$. – Viktor Vaughn Jun 11 '19 at 07:55
  • Right, it seems I assumed a wrong definition of exactness of sheaves. – George Jun 11 '19 at 12:43