Exact functor on abelian categories

Let $𝖢, 𝖣$ be abelian categories and $𝐹 : 𝖢 \to 𝖣$ be an $𝖠 𝖻$ -functor. Then #m/thm/homology

$𝐹$ is left exact iff it preserves kernels; equivalently for any exact sequence $0 \to 𝑋 \to 𝑌 \to 𝑍$ the sequence $0 \to 𝐹 𝑋 \to 𝐹 𝑌 \to 𝐹 𝑍$ is exact.
$𝐹$ is right exact iff it preserves cokernels; equivalently for any exact sequence $𝑋 \to 𝑌 \to 𝑍 \to 0$ the sequence $𝐹 𝑋 \to 𝐹 𝑌 \to 𝐹 𝑍 \to 0$ is exact.

Thus $𝐹$ is exact iff it preserves short exact sequences.

Proof

It suffices to show the left exact case, whence the right exact case follows by duality.

Suppose $𝐹$ is left exact and $0 \to 𝑋 \to 𝑌 \to 𝑍$ is an exact sequence. Then the designated arrows $0 \to 𝑋$ and $𝑋 \to 𝑌$ are the kernels of $𝑋 \to 𝑌$ and $𝑌 \to 𝑍$ respectively. It follows $0 \to 𝐹 𝑋$ and $𝐹 𝑋 \to 𝐹 𝑌$ are the kernels of $𝐹 𝑋 \to 𝐹 𝑌$ and $𝐹 𝑌 \to 𝐹 𝑍$ respectively, so the sequence $0 \to 𝐹 𝑋 \to 𝐹 𝑌 \to 𝐹 𝑍$ is exact.

For the converse, if for any exact $0 \to 𝑋 \to 𝑌 \to 𝑍$ we have $0 \to 𝐹 𝑋 \to 𝐹 𝑌 \to 𝐹 𝑍$ exact, then $𝐹$ preserves kernels. It must also preserve biproducts since it is $𝖠 𝖻$ -enriched. Thus, by the limit construction theorems we have a left exact functor.

Exact functor on abelian categories

See also