About an embedding of abelian categories into categories of modules

Question

Let $k$ be a field. Let $C$ be an abelian $k$-linear category with a symmetric tensor product $\otimes$ and internal homomorphisms, such that $\mathrm{End}(1)=k$. Let $M$ be another $k$-linear abelian category, and let $$h:M\to C$$ be an exact, faithful, $k$-linear functor. If $C$ has arbitrary limits and $M$ is small, we can understand the endomorphism ring of $h$ as a ring object (a monoid) in $C$, as follows. It is the equaliser $$R := \ker \left( \prod_{m\in M}\mathrm{End}(h(m)) \xrightarrow{\quad u \quad} \prod_{m\to n}\mathrm{Hom}(h(m),h(n))\right)$$ where $u$ is "precomposition minus postcomposition". The given functor $h$ factors then canonically over the category of $R$-modules in $C$ $$M \xrightarrow{\quad\widetilde h \quad} R\mathrm{-Mod}_C \xrightarrow{\quad f \quad} C$$ where $f$ is the forgetful functor. My question is:

Is the functor $\widetilde h$ fully faithful?

For example, if $C$ is the category of vector spaces over $k$, then the answer is Yes. For this, $k$ does not even have to be a field. On the other hand, $C$ could be much larger, for example, the category of sheaves of $k$-vector spaces on a connected topological space. This is the case i am ultimately interested in.

Remark: The existence of limits in $C$ was just for convenience, without them, one can still view $R$ as a pro-ring object and consider $R$-modules. The hypothesis $\mathrm{End}(1)=k$ is essential if one wishes for a positive answer, because the category of modules $R\mathrm{-Mod}_C$ is always $\mathrm{End}(1)$-linear.

Remark: Also, the whole story smells like it was some variant of the Freyd-Mitchell theorem, but i don't see a concrete way to link it to that.

Added: The proof that in the case $C = \mathrm{Vect}(k)$ the answer to my question is Yes goes roughly as follows: In this special case, we can view $M$, or at least the $\mathrm{Ind}M$, as a $C$-module, that is, for every vector space $V$ and object $m \in M$ we can give a sense to $\mathrm{Hom}(V,m)$ and $V\otimes m$ as objects in $M$. At this point, $\widetilde h$ is then even an equivalence. Define an object $$X := \ker \left( \prod_{m\in M}\mathrm{Hom}(h(m),m) \xrightarrow{\quad u \quad} \prod_{m\to n}\mathrm{Hom}(h(m), n)\right)$$ in $M$. Then $h(X)$ is $R$ as a right $R$-module, and given a left $R$-module $V$, we can define an object $Y$ of $M$ by $$Y = X \otimes_R V = \mathrm{coker}(X \otimes R \otimes V \xrightarrow{\quad v \quad} X \otimes V)$$ with $v$ = "left action on $V$ minus right action $X$". But then, $\widetilde h(Y)$ is isomorphic to the given $R$-module $V$, and one can treat morphisms in the same way: $V\to V'$ yields $X\otimes_RV \to X\otimes_RV'$. Note that if we only consider finite dimensional spaces, then the passage to ind-objects is superfluous, and $\widetilde h$ is an equivalence of categories.

The proof generalises for as far as $M$ is a $C$-module. But what is there to do if, for example, $M$ is "graded local systems" on a topological space and $C$="sheaves" and $h$=forget?

Answer 1

No. What follows appears to be a counterexample for $C = \text{Vect}$ (I don't understand where in your argument you prove fullness).

Let $M = \text{Vect}^{op}, C = \text{Vect}$, and let $h : \text{Vect}^{op} \to \text{Vect}$ be the contravariant functor $V \mapsto V^{\ast} \cong \text{Hom}(V, k)$. If smallness is important to you pretend that the first $\text{Vect}$ means vector spaces of at most countable dimension. The endomorphism ring of $h$ is $k$, so the lift $\tilde{h}$ is just $h$ again.

I claim that $h$ is not full. To see this, if $V$ is a countable-dimensional vector space, regarded as an object in $\text{Vect}^{op}$, then the induced map

$$\text{End}(V, V) \to \text{End}(V^{\ast}, V^{\ast})$$

has the property that its source has dimension $\aleph_0^{\aleph_0} = 2^{\aleph_0}$, but its target has dimension at least $\left( 2^{\aleph_0} \right)^{2^{\aleph_0}}$.

(Taking the opposites of familiar abelian categories seems to be my new favorite trick! Note that vector space duality establishes that $\text{FinVect}$ is equivalent to its opposite. $\text{Vect}$ itself is the ind-completion of $\text{FinVect}$, so $\text{Vect}^{op}$ is the pro-completion; in other words, it's the category of profinite vector spaces. This category is also known as the category of linearly compact vector spaces: see this MO question and this n-cafe post for details. This gives some intuition for why $h$ is not full: it's the forgetful functor and it doesn't see the topology.

In the special case that $k = \mathbb{F}_p$ we can be a little more explicit: $\text{Vect}^{op}$ in this case is the full subcategory of profinite groups consisting of the ones whose finite quotients are all elementary abelian $p$-groups.)