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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.)

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.)

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 provector 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.)

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.)

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 for $k$ sufficiently large (I think uncountable suffices). 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 $\aleph_0^{\aleph_0}$, but its target has dimension at least $\left( \aleph_0^{\aleph_0} \right)^{\aleph_0^{\aleph_0}} = \aleph_0^{\aleph_0^{\aleph_0}}$.

(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 provector 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.)

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 provector 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.)