$\DeclareMathOperator{\Hom}{Hom}$ $\DeclareMathOperator{\Mod}{Mod}$ $\DeclareMathOperator{\End}{End}$ $\DeclareMathOperator{\coker}{coker}$I'm going through the proof of the following theorem :
Let $k$ be a commutative ring (with unity). Let $C$ be a $k$-abelian category and $R$ a $k$-algebra. Assume that $C$ admits small inductive limits. Then for any $X\in \Mod(R,C)$ and $N\in \Mod(R^{op})$ the functor $Y\mapsto \Hom_{R^{op}}(N,\Hom_C(X,Y))$ is representable. Furthermore, denoting $N\otimes_R X$ its representative, the functor $$\bullet \otimes_R \bullet : \Mod(R^{op})\times \Mod(R,C)\to C $$ is additive and right exact in each variable.
I'm not sure how standard the definitions are, so to be clear, $k$-Abelian category is the same as an abelian category but $\Hom_C(X,Y)\in \Mod(k)$ (instead of $\Mod(\mathbb{Z}))$, with composition being $k$-bilinear. Also, $\Mod(R,C)$ is defined as follows :
- The objects are pairs $(X,\alpha_X)$ where $X$ is an object in $C$ and $\alpha_X:R\to \End_C(X)$ is a $k$-algebra morphism.
- A morphism $(X,\alpha_X)\to(Y,\alpha_Y)$ is a morphism $f:X\to Y$ in $C$ such that $\alpha_Y\circ f = f\circ\alpha_X$
When $C$ is $k$-abelian, $\Mod(R,C)$ is as well, and the forgetful functor $\Mod(R,C)\to C$ is faithful and exact.
I'm having troubles with details in the proof. First the author proves that when $N=R^{\oplus I}$ for a small set $I$, the assumption implies the following chain of isomorphism, and they are functorial in $Y$ : $$ \Hom_R(R^{\oplus I},\Hom_C(X,Y))\simeq \Hom_C(X,Y)^I\simeq \Hom_C(X^{\oplus I}, Y) $$
I'm not sure about what follows : as I understand the first isomorphism is the universal property of the sum (coproduct) i.e. $\Hom(R^{\oplus I},B)\simeq \Hom(R,B)^{\oplus I}$ but $R$ being free, $\Hom(R,B)\simeq B$. Then it's again the universal property of the sum, and using that $C$ admits small colimits, but I'm honestly not sure where the assumption is used.
Then the author proceeds to treat the general case. Starting with the following fact : For all $N\in \Mod(R^{op})$ we can find an exact sequence $$ R^{\oplus J}\to R^{\oplus I}\to N\to 0 $$ with $I,J$ small sets. I can see why one can find $R^{\oplus I}\to N\to 0$, say taking the free module on the set $N$ we do get a surjective map as required. What I don't know, is how can we be sure to find $R^{\oplus J}\to R^{\oplus I}$ making the sequence exact. If $R$ was a PID then submodule of free module are again free, so the kernel $R^{\oplus I}\to N$ would be fine, but this is not true in general so I'm not sure.
Finally, from the exact sequence stated above, applying the left exact functor $\Hom_R(\bullet, \Hom_C(X,Y))$ we get the exact sequence : $$ 0\to \Hom_R(N,\Hom_C(X,Y))\to \Hom_R(R^{\oplus I},\Hom_C(X,Y))\to \Hom_R(R^{\oplus J},\Hom_C(X,Y)) $$ The author concludes that $\coker(X^{\oplus J}\to X^{\oplus I})$ represents $N\otimes_R X$. I don't quite understand the last step here, how does one conclude that ?
Thanks for any help or hint.
As far as I understand the question, you have three main questions.
Firstly, note that a $k$-algebra $R$ is just a single object small $k$-linear abelian category. The category $Mod(R, C)$ (called the category of left $R$-objects in $C$) is just the full subcategory of the functor category $[R,C]$ consisting of all $k$-linear functors. In particular, $Mod(R^{op}) = Mod(R^{op}, k-Mod)$ is the category of right-$R$-modules.
Question 1: Note that the contravariant representable hom on any category $A$, namely $Hom(-,a) : A^{op} \rightarrow Set$(or $k-Mod$ etc.) changes colimits into limits. In particular, it turns coproducts into products. Thus, the correct statement is : $Hom_R (R^{\oplus I}, B) \cong \prod_I Hom_R (R, B) \cong B^I$. This is what is used in the first isomorphism. The second isomorphism also uses the same result, but now applied to the contravariant hom, $Hom_C (-, Y)$.
Question 2 : This is completely standard. Take any right $R$-module $N$. As you mentioned, getting a surjection $R^{\oplus I} \rightarrow N$ is easy, using the universal property of free modules. Look at the kernel of this surjection. You get a surjection $R^{\oplus J} \rightarrow ker(R^{\oplus I} \rightarrow N)$. Thus, you have a right exact sequence $R^{\oplus J} \rightarrow R^{\oplus I} \rightarrow N \rightarrow 0$.
Question 3: You have a left exact sequence $0 \rightarrow Hom_R(N, Hom_C(X,Y)) \rightarrow Hom_R(R^{\oplus I}, Hom_C(X,Y)) \rightarrow Hom_R(R^{\oplus J}, Hom_C(X,Y))$. Further step 1 gives you a commutative diagram :
$\require{AMScd}$ \begin{CD} 0 @>>> Hom_R(N, Hom_C(X,Y)) @>>> Hom_R(R^{\oplus I}, Hom_C(X,Y)) @>>> Hom_R(R^{\oplus J}, Hom_C(X,Y))\\ @. @. @V{\cong}VV @V{\cong}VV\\ @. @. Hom_C(X^{\oplus I},Y)@>>> Hom_C(X^{\oplus J}, Y) \end{CD} where the bottom horizontal arrow is induced by the fact that the vertical arrows are isomorphisms. Now, the bottom horizontal arrow is natural in the variable $Y$ (by construction), so that what you have is a natural transformation $Hom_C(X^{\oplus I}, -) \Rightarrow Hom_C(X^{\oplus J}, -)$. By the Yoneda lemma, this is induced by an arrow $X^{\oplus J} \rightarrow X^{\oplus I}$. Take the cokernel of this arrow, to get a right exact sequence $X^{\oplus J} \rightarrow X^{\oplus I} \rightarrow Coker \rightarrow 0$.
Apply $Hom(-,Y)$ on this, you now get a commutative diagram :
$\require{AMScd}$ \begin{CD} 0 @>>> Hom_R(N, Hom_C(X,Y)) @>>> Hom_R(R^{\oplus I}, Hom_C(X,Y)) @>>> Hom_R(R^{\oplus J}, Hom_C(X,Y))\\ @. @VVV @V{\cong}VV @V{\cong}VV\\ 0 @>>> Hom_C(Coker, Y) @>>> Hom_C(X^{\oplus I},Y)@>>> Hom_C(X^{\oplus J}, Y) \end{CD} where the vertical arrow $Hom_R(N, Hom_C(X,Y)) \rightarrow Hom_C(Coker, Y)$ is induced by the universal property of $ker (Hom_C(X^{\oplus I}, Y)) \rightarrow Hom_C(X^{\oplus J}, Y))) = Hom_C(Coker, Y)$. The five lemma guarantees that this vertical arrow is indeed an isomorphism. Again this vertical isomorphism $Hom_R(N, Hom_C(X,Y)) \xrightarrow{\sim} Hom_C(Coker, Y))$ is natural in $Y$, so that $Hom_R(N, Hom_C(X,-)) \cong Hom_C(Coker,-)$ and therefore by the Yoneda lemma, $N \otimes_R X \cong Coker(X^{\oplus J} \rightarrow X^{\oplus I})$.
Hope this clears up your doubts.