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](https://math.stackexchange.com/questions/2864242/is-the-class-of-dualizable-objects-in-an-abelian-monoidal-category-closed-under), which however didn't receive an answer up to know.



Tannaka-Krein reconstruction and rigidity
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).


