$\DeclareMathOperator{\Mod}{Mod}$ $\DeclareMathOperator{\Coim}{Coim}$ $\DeclareMathOperator{\Img}{Im}$ $\DeclareMathOperator{\End}{End}$ I'm currently going through some parts of a book called "Categories and Sheaves". I'm trying to spell out some details said to be obvious but I'm having slight issues. For completeness' sake, I'll state the definition I'm using here, which are the ones from that book.
A category $C$ is additive if for all $X,Y\in C$ $Hom_C(X,Y)\in \Mod(\mathbb{Z})$, and $\circ$ is bilinear, and $C$ has a zero object, and $C$ has finite products and coproducts.
A category $C$ is abelian if it is additive and every morphism has a kernel and a cokernel, and for all morphism $f$ in $C$ the natural morphism $u: \Coim f \to \Img f$ is an isomorphism.
Let $k$ be an commutative ring with identity, we say $C$ is $k$-additive if we replace $\Mod(\mathbb{Z})$ in the definition above with $\Mod(k)$ and bilinear with $k$-bilinear. We also obtain the notion of $k$-abelian category.
Finally, let $C$ be a $k$-additive category, $R$ a $k$-algebra, we define the category $\Mod(R,C)$ by :
- It's objects are pairs $(X,\alpha_X)$ where $X\in C$ and $\alpha_X:R\to \End_C(X) $ is a morphism of $k$-algebras.
- A morphism $(X,\alpha_X)\to (Y,\alpha_Y)$ is a morphism $f:X\to Y$ in $C$ such that $f\circ \alpha_X(r) = \alpha_Y(r)\circ f$ for all $r\in R$.
$C$ being $k$-additive, $\Mod(R,C)$ is also $k$-additive. The zero object being $(0,0:R\to \End_C(X))$, the $k$-module structure being the obvious one, and similarly the bilinearity is direct. Again, the existence of finite products and coproducts I can prove.
1 / The next statement is where I start having issues. The authors state that if $F:C\to C'$ is a $k$-additive functor, it induces a functor $F_R: \Mod(R,C)\to \Mod(R,C')$. I sort of see why that should be the case but I tried writing it out and I'm not sure how exactly $F_R$ acts on the objets, that is $F_R(X,\alpha_X)=(FX,?)$, I'm confused at what the morphism of $k$-algebras $R\to \End_{C'}(FX)$ should be, given $F$ and $\alpha_X$.
2 / Then the authors move on to the following proposition
Let $C$ be a $k$-abelian category, then $\Mod(R,C)$ is $k$-abelian and the natural forgetful functor is faithful and exact, and if $F:C\to C'$ is a functor of $k$-abelian categories and right (rest. left) exact, $F_R$ is right (resp. left) exact.
The proof is supposedly obvious, but I have an issue of the same style as before, to define the kernel of $f:(X,\alpha_X)\to (Y,\alpha_Y)$ I'd want an object $(K,\alpha_K)$ mapping to $(X, \alpha_X)$. It seems clear that $K$ should be the object $\ker f$ in C, but I'm not sure what $\alpha_{\ker f}$ would be. Of course the same question arises when thinking about the cokernel of a map in $\Mod(R,C)$.
$\DeclareMathOperator{\End}{End}$I was able to solve my problems and it was indeed quite easy.
First as pointed out in the comments $F_R(X,\alpha_X)=(FX,F\circ \alpha_X)$, this makes sense, because $\alpha_X:R\to \End_C(X)$ hence $F(\alpha_X(r))\in \End_{C'}(FX)$ for every $r\in R$, it all works nicely as expected.
Second, given $f:(X,\alpha_X)\to (Y,\alpha_Y)$ the question was, what is the kernel. Well say (if it exists) it's $(K,\alpha_K)$,then you'd want $K$ to be the kernel of $f$ seen as a map in $C$. But then $\alpha_K(r)$ should be a morphism $K\to K$ which is the same as a morphism $K\to X$ such that the composition $K\to X \xrightarrow{f} Y$ is zero. For $r\in R$ let $t(r)=\alpha_X(r)\circ\ker f:K\to X$. We have $f\circ t(r) = f\circ \alpha_X(r) \circ \ker f = \alpha_Y(r)\circ f \circ \ker f = \alpha_Y(r) \circ 0 = 0$ hence $t$ factors through $K$ as we wanted. We can then take $\alpha_K$ to be the induced map $K\to K$, showing then that $(K,\alpha_K)$ is the kernel of the map $f$ in $\operatorname{Mod}(R,C)$ is straightforward.