Tannaka-Krein reconstruction and rigidity Let $\mathcal{C}$ be a rigid monoidal category together with a quasi-monoidal functor $\omega:\mathcal{C}\to\mathsf{vec}_{\Bbbk}$ to finite-dimensional vector spaces over a field $\Bbbk$, i.e. we have isomorphisms $\varphi_0:\Bbbk\to\omega(\mathbb{I})$ and $\varphi=\left(\varphi_{X,X}:\omega(X)\otimes \omega(Y)\to\omega(X\otimes Y)\right)_{X,Y\in\mathcal{C}}$ but these are not necessarily compatible with the constraints.
In light of Majid's Tannaka-Krein theorem for quasi-Hopf algebras and other results we know that there exists a coquasi-bialgebra $H$ (namely, $H=\mathsf{coend}(\omega)$) such that $\omega$ factors through the forgetful functor $\mathcal{U}:\mathsf{Com}^{H}_{fd}\to\mathsf{vec}_{\Bbbk}$ from the category of finite-dimensional $H$-comodules to $\mathsf{vec}_{\Bbbk}$, that is to say, there exists a functor $\mathcal{I}_{\mathcal{C}}:\mathcal{C}\to \mathsf{Com}^{H}_{fd}$ such that $\mathcal{U}\mathcal{I}_{\mathcal{C}} = \omega$.
I have been told that:

Claim: Since any object in $\mathsf{Com}^{H}_{fd}$ can be obtained from objects in the image of $\mathcal{I}_{\mathcal{C}}$ by taking direct sums, kernel and cokernels and since in an abelian monoidal category with exact tensor product (as $\mathsf{Com}^{H}_{fd}$) the collection of rigid objects is closed under taking sums, kernels and cokernels, the category $\mathsf{Com}^{H}_{fd}$ has to be rigid as well.

My main question is:

Main Question: Is this true?

It is since a long time that I am trying to prove this claim, but I couldn't manage to.
In light of Schauenburg's Tannaka duality for arbitrary Hopf algebras, Corollary 2.2.9, every object in $\mathsf{Com}^{H}_{fd}$ is the quotient of a subobject of a finite biproduct of objects from the image of the functor $\mathcal{I}_{\mathcal{C}}$, but this is not the same as Claim, whence I provided my own proof of the fact that any object in $\mathsf{Com}^{H}_{fd}$ can be obtained from objects in the image of $\mathcal{I}_{\mathcal{C}}$ by taking direct sums, kernel and cokernels. It is quite long and technical, but I think it should work, whence I'll omit it.
My problem now is with the second part of the Claim, about abelian monoidal categories. Since I didn't manage to deal with it directly, I am trying to approach the problem step by step.
Let $\Bbbk$ be a von Neumann regular commutative algebra (i.e. every $\Bbbk$-module is flat) and let $\mathsf{Mod}_{\Bbbk}$ be the category of $\Bbbk$-modules. The collection of rigid objects here should coincide with the finitely-generated and projective modules. Thus, if $f:M\to N$ is a morphism of fgp $\Bbbk$-modules:

Sub-question 1: is it true that $\ker(f)$ is fgp as well?
Sub-question 2: what about $\operatorname{coker}(f)$?

Any help or comment (even rude ones) will be very welcome.
NB: this question is strictly related with this other question on MSE, which however didn't receive an answer up to know.
 A: It seems that asking helps in enlightening. Let $\mathcal{C}$ be an abelian monoidal category with exact tensor product and let $f:X\rightarrow Y$ be a morphism between (right) rigid objects in $\mathcal{C}$. We want to prove that $\ker \left( f\right) ^{{\star }}=\mathrm{coker}\left( f^{{\star }}\right) $ and that $\mathrm{coker}\left( f\right)^{{\star }}=\ker \left( f^{{\star }}\right) $. Let us adopt the following notation
\begin{gather*}
0 \longrightarrow \ker \left( f\right) \overset{k}{\longrightarrow }X\overset{f}{\longrightarrow }Y\overset{c}{\longrightarrow }\mathrm{coker}\left( f\right) \longrightarrow 0, \\
0 \longrightarrow \ker \left( f^{{\star }}\right) \overset{k_{{\star }}}{\longrightarrow }Y^{{\star }}\overset{f^{{\star }}}{\longrightarrow }X^{{\star }}\overset{c_{{\star }}}{\longrightarrow }\mathrm{coker}\left(
f{^{\star }}\right) \longrightarrow 0.
\end{gather*}
First of all, we define $\mathsf{db}_{k}:\boldsymbol{1}\rightarrow \mathrm{coker}\left( f\right) ^{{\star }}\otimes \ker \left( f\right) $ as follows.
Consider the composition
$$
\boldsymbol{1}\overset{\mathsf{db}_{X}}{\longrightarrow }X{^{\star
}}\otimes X\overset{c_{{\star }}\otimes X}{\longrightarrow }\mathrm{coker}
\left( f^{{\star }}\right) \otimes X.
$$
We have that
\begin{align*}
\left( \mathrm{coker}\left( f^{{\star }}\right) \otimes f\right) \circ
\left( c_{{\star }}\otimes X\right) \circ \mathsf{db}_{X} &=\left(
c_{{\star }}\otimes Y\right) \circ \left( X^{{\star }}\otimes f\right)
\circ \mathsf{db}_{X} \\
&=\left( c_{{\star }}\otimes Y\right) \circ \left( f{^{\star }}\otimes
Y\right) \circ \mathsf{db}_{Y}=0
\end{align*}
and hence $\left( c_{{\star }}\otimes X\right) \circ \mathsf{db}_{X}$
factors through the kernel of $\mathrm{coker}\left( f{^{\star }}\right)
\otimes f$, i.e. there exists a unique $\mathsf{db}_{k}$ such that 
$$
\left( c_{{\star }}\otimes X\right) \circ \mathsf{db}_{X}=\left( \mathrm{
coker}\left( f^{{\star }}\right) \otimes k\right) \circ \mathsf{db}_{k}.
$$
Secondly, we define $\mathsf{ev}_{k}:\ker \left( f\right) \otimes \mathrm{
coker}\left( f^{{\star }}\right) \rightarrow \boldsymbol{1}$ as follows.
Consider the composition
$$
\ker \left( f\right) \otimes X^{{\star }}\overset{k\otimes X^{{\star }}}{
\longrightarrow }X\otimes X^{{\star }}\overset{\mathsf{ev}_{X}}{
\longrightarrow }\boldsymbol{1}.
$$
We have that
\begin{align*}
\mathsf{ev}_{X}\circ \left( k\otimes X^{{\star }}\right) \circ \left( \ker
\left( f\right) \otimes f^{{\star }}\right)  &=\mathsf{ev}_{X}\circ \left(
X\otimes f^{{\star }}\right) \circ \left( k\otimes Y^{{\star }}\right)  \\
&=\mathsf{ev}_{Y}\circ \left( f\otimes Y^{{\star }}\right) \circ \left(
k\otimes Y^{{\star }}\right) =0
\end{align*}
and hence $\mathsf{ev}_{X}\circ \left( k\otimes X{^{\star }}\right) $
factors through the cokernel of $\ker \left( f\right) \otimes f{^{\star }}$
, i.e. there exists a unique $\mathsf{ev}_{k}$ such that
$$
\mathsf{ev}_{k}\circ \left( \ker \left( f\right) \otimes c_{{\star
}}\right) =\mathsf{ev}_{X}\circ \left( k\otimes X{^{\star }}\right) .
$$
Let us check that these satisfy the zigzag identities. We compute
\begin{align*}
k &\circ \left( \mathsf{ev}_{k}\otimes \ker \left( f\right) \right) \circ
\left( \ker \left( f\right) \otimes \mathsf{db}_{k}\right)  \\
&=\left( \mathsf{ev}_{k}\otimes X\right) \circ \left( \ker \left( f\right) \otimes \mathrm{coker}\left( f^{{\star }}\right) \otimes k\right) \circ \left( \ker\left( f\right) \otimes \mathsf{db}_{k}\right)  \\
&=\left( \mathsf{ev}_{k}\otimes X\right) \circ \left( \ker \left( f\right)
\otimes c_{{\star }}\otimes X\right) \circ \left( \ker \left( f\right)
\otimes \mathsf{db}_{X}\right)  \\
&=\left( \mathsf{ev}_{X}\otimes X\right) \circ \left( k\otimes X{^{\star
}}\otimes X\right) \circ \left( \ker \left( f\right) \otimes \mathsf{db}
_{X}\right)  \\
&=\left( \mathsf{ev}_{X}\otimes X\right) \circ \left( X\otimes \mathsf{db}
_{X}\right) \circ k=k,
\end{align*}
from which we deduce (since $k$ is mono) that $\left( \mathsf{ev}_{k}\otimes
\ker \left( f\right) \right) \circ \left( \ker \left( f\right) \otimes 
\mathsf{db}_{k}\right) =\mathrm{id}_{\ker \left( f\right) }$, and
\begin{align*}
\left( \mathrm{coker}\left( f^{{\star }}\right) \otimes \mathsf{ev}
_{k}\right) & \circ \left( \mathsf{db}_{k}\otimes \mathrm{coker}\left(
f^{{\star }}\right) \right) \circ c_{{\star }} \\
&=\left( \mathrm{coker}\left( f^{{\star }}\right) \otimes \mathsf{ev}_{k}\right) \circ \left( \mathrm{coker}\left( f^{{\star }}\right) \otimes \ker \left( f\right)\otimes c_{{\star }}\right) \circ \left( \mathsf{db}_{k}\otimes X^{{\star}}\right)  \\
&=\left( \mathrm{coker}\left( f^{{\star }}\right) \otimes \mathsf{ev}
_{X}\right) \circ \left( \mathrm{coker}\left( f^{{\star }}\right) \otimes
k\otimes X^{{\star }}\right) \circ \left( \mathsf{db}_{k}\otimes X^{{\star
}}\right)  \\
&=\left( \mathrm{coker}\left( f{^{\star }}\right) \otimes \mathsf{ev}
_{X}\right) \circ \left( c_{{\star }}\otimes X\otimes X^{{\star }}\right)
\circ \left( \mathsf{db}_{X}\otimes X^{{\star }}\right)  \\
&=c_{{\star }}\circ \left( X^\star\otimes \mathsf{ev}_{X}\right) \circ \left( 
\mathsf{db}_{X}\otimes X^{{\star }}\right) =c_{{\star }},
\end{align*}
from which we deduce that $\left( \mathrm{coker}\left( f{^{\star }}\right)
\otimes \mathsf{ev}_{k}\right) \circ \left( \mathsf{db}_{k}\otimes \mathrm{
coker}\left( f^{{\star }}\right) \right) =\mathrm{id}_{\mathrm{coker}\left(
f^{{\star }}\right) }$.
The other one is analogous and hence the claim holds (in particular, also for modules over a von Neumann regular ring).
