Let $F \colon \mathcal{A} \to \mathcal{B}$ an exact functor between abelian categories, $X^\bullet$ and $Y^\bullet$ chain complexes and $f \colon X^\bullet \to Y^\bullet$ a morphism. I have to prove that $F(C(f))$ is isomorphic to $C(F(f))$, where $C(f)$ is the (mapping) cone of $f$.
I know that, because $F$ is additive, it preserves biproduct, so $F(X^\bullet \oplus Y^\bullet) \cong F(X^\bullet) \oplus F(Y^\bullet)$
I have some problems about the differentials.
The differential of $C(f)$ is $\forall n \in \mathbb{Z}$ $d_{C(f)_n} = \begin{bmatrix} -d_{X_{n-1}} & 0 \\ f_{n-1} & d_{Y_n}\end{bmatrix}$.
The differential of $F(C(f))$ is $\forall n \in \mathbb{Z}$ $d_{F(C(f))_n} = \begin{bmatrix} -F(d_{X_{n-1}}) & 0 \\ F(f_{n-1}) & F(d_{Y_n}) \end{bmatrix}$.
How can I prove: $F(d_{X_{n-1}}) = d_{F(X_{n-1})}$?