Suppose $F: C\to D$ is an additive functor between abelian categories and that

$$0\to X\xrightarrow f Y\xrightarrow g Z\to 0$$

is and exact sequence in $C$. Does it follow that $F(X)\xrightarrow{F(f)} F(Y)\xrightarrow{F(g)} F(Z)$ is exact in $D$? In other words, is $\ker(F(g))=\mathrm{coker}(F(f))$?

Remark 1: If the answer is no, a counterexample must use a non-split short exact sequence. This is because additive functors send split exact sequences to split exact sequences. A splitting is a pair $s:Y\to X$ and $r:Z\to Y$ so that $id_Y=f\circ s+r\circ g$, $id_X=s\circ f$, and $id_Z=g\circ r$. An additive functor preserves these properties, so $F(s)$ and $F(r)$ will split the sequence in $D$.

Remark 2: You probably know you know lots of left exact and right exact additive functors, but you also know lots of exact in the middle additive functors. $H^i$ and $H_i$ for any (co)homology theory are neither left or right exact, but they are exact in the middle by the long exact sequence in (co)homology.

Suppose $F: C\to D$ is an additive functor between abelian categories and that

$$0\to X\xrightarrow f Y\xrightarrow g Z\to 0$$

is and exact sequence in $C$. Does it follow that $F(X)\xrightarrow{F(f)} F(Y)\xrightarrow{F(g)} F(Z)$ is exact in $D$? In other words, is $\ker(F(g))=\mathrm{im}(F(f))$?

Remark 1: If the answer is no, a counterexample must use a non-split short exact sequence. This is because additive functors send split exact sequences to split exact sequences. A splitting is a pair $s:Y\to X$ and $r:Z\to Y$ so that $id_Y=f\circ s+r\circ g$, $id_X=s\circ f$, and $id_Z=g\circ r$. An additive functor preserves these properties, so $F(s)$ and $F(r)$ will split the sequence in $D$.

Remark 2: You probably know you know lots of left exact and right exact additive functors, but you also know lots of exact in the middle additive functors. $H^i$ and $H_i$ for any (co)homology theory are neither left or right exact, but they are exact in the middle by the long exact sequence in (co)homology.