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Suppose $$0 \longrightarrow A \longrightarrow B\longrightarrow C \longrightarrow 0$$ is a short exact sequence and $F$ is a functor, so we get maps $F(A)\longrightarrow F(B)$ and $F(B) \longrightarrow F(C)$. Is the sequence $$F(A)\longrightarrow F(B)\longrightarrow F(C)$$ exact in general?

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No. Why would we bother with the name "half-exact functor" if it was automatic? – Zhen Lin Nov 21 '13 at 22:01

You can find a couple of examples at MO/46019.

Note that since you didn't assume that $F$ is additive, there are lots of trivial examples (for example non-zero constant functors).

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