Let me set the notations:
- $(\mathcal{A}, T)$ will be an additive category with translation
- $T(X)$ will denote the application of the functor $T$ to an object $X \in \mathcal{A}$
- A differential object in $\mathcal{A}$ is an object $X$ together with a morphism $d_X : X \rightarrow T(X)$
- TX will denote the shifted object of a differential object: $T(X) \rightarrow T^2(X)$, where the differential is $-T(d_X)$
- For a morphism of differential objects $f: X \rightarrow Y$ I will denote $Mc(f)$ the cone of $f$
Let now $F: (\mathcal{C},T) \rightarrow (\mathcal{D},T')$ be an additive functor between additive categories with translation. Prove that $F(Mc(f)) \cong Mc(F(f))$.
What I did so far:
- $F$ additive $\implies$ F commutes with direct sums
- $F(TX) \cong T'F(X)$
The problem I am dealing with is that when constructing the complexes $F(-)$ for some differential object I must take into account the isomorphism $F \circ T \cong T' \circ F$, hence it becomes difficult to cope with the notations. Is there a more elegant way to prove that statement or is just a matter of writing down the required maps and then everything follows?