# 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.

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 $\left( \mathrm{coker}\left( f{^{\star }}\right) \otimes \mathsf{ev}_{k}\right) \circ \left( \mathsf{db}_{k}\otimes \mathrm{ coker}\left( f^{{\star }}\right) \right) =\mathrm{id}_{\mathrm{coker}\left( f^{{\star }}\right) }$.