# Is the class of dualizable objects in an abelian monoidal category closed under sums, kernels and cokernels?

Goodmorning to everybody.

I am in the following situation.

I have been told that in an abelian monoidal category (I assume this means an abelian category $\mathscr{A}$ with a monoidal structure $(\mathscr{A},\otimes,\mathbb{I},\alpha,\lambda,\rho)$ such that the tensor product $\otimes$ is linear in each component) with exact tensor product the collection of (right) rigid (or dualizable) objects (i.e. those $X$ in $\mathscr{A}$ for which there exist $X^*$, $\mathsf{ev}_X:X\otimes X^*\to\mathbb{I}$ and $\mathsf{coev}_X:\mathbb{I}\to X^*\otimes X$ in $\mathscr{A}$ as well, satisfying the triangle or zigzag identities) is closed under sums (I imagine, biproducts, which I am going to denote by $\oplus$), kernels and cokernels. I am trying to verify this claim, which has been presented to me as trivial.

I know that since $\otimes$ is linear in each variable, it distributes over $\oplus$ as in the obvious example $\mathscr{A}=\mathsf{Vec}_{\Bbbk}$. Therefore, if $X,Y$ in $\mathscr{A}$ are rigid, I can consider the sum $X^*\oplus Y^*$, the morphism $\mathsf{ev}_{X\oplus Y}:(X\oplus Y)\otimes (X^*\oplus Y^*)\to\mathbb{I}$ given (up to isomorphism) by the coproduct of the arrows $\mathsf{ev}_X:X\otimes X^*\to\mathbb{I}, \mathsf{ev}_Y:Y\otimes Y^*\to\mathbb{I}, \boldsymbol{0}:X\otimes Y^*\to\mathbb{I}, \boldsymbol{0}:Y\otimes X^*\to \mathbb{I}$, and the morphism $\mathsf{coev}_{X\oplus Y}:\mathbb{I}\to (X^*\oplus Y^*)\otimes (X\oplus Y)$ given (up to isomorphism) by the product of the arrows $\mathsf{coev}_X:\mathbb{I}\to X^*\otimes X$, $\mathsf{coev}_Y:\mathbb{I}\to Y^*\otimes Y$, $\boldsymbol{0}:\mathbb{I}\to X^*\otimes Y$, $\boldsymbol{0}:\mathbb{I}\to Y^*\otimes X$. I didn't check the details, but I may trust that these give a dual object for $X\oplus Y$.

For kernels and cokernels I didn't find a way to approach the question. I have been told, as a hint, that for a morphism $f:X\to Y$ between dualizable objects in $\mathscr{A}$, the tensor product with the kernel of $f$ (resp. $f^*$) is right adjoint to the tensor product with the cokernel of $f^*$ (resp. $f$).

Here come my questions:

1. Is it really the question so trivial, that my problem is simply that I am looking at it from the wrong view-point?
2. Is it true, in an abelian monoidal category, that if $-\otimes X$ admits $-\otimes Y$ as a left adjoint, then $X$ has to be rigid and $Y\cong X^*$?
3. How can one show that tensoring by the kernel of $f$ is right adjoint to tensoring by the cokernel of the dual map?
4. Is there a particular reference where these matters about rigid objects in abelian monoidal categories are treated?

About point 2., I guess that if $\eta:(-)\to (-)\otimes Y\otimes X$ and $\epsilon:(-)\otimes X\otimes Y\to (-)$ are the unit and the counit of the adjunction, then $\mathsf{ev}_X = \epsilon_{\mathbb{I}}$ and $\mathsf{coev}_X=\eta_{\mathbb{I}}$, but then I am in trouble in showing that $(\epsilon_{\mathbb{I}}\otimes X) \circ \alpha_{X,Y,X}^{-1} \circ (X\otimes \eta_{\mathbb{I}}) = \mathsf{id}_X$, for example.

Any help or comment (even rude ones) would be very welcome.

[Edit 27.08.2018] Since it seems that the question is too complicated to face directly, I decided to try to look for a "concrete" example. In this direction, let $\Bbbk$ be a von Neumann regular ring (this means that every $\Bbbk$-module is flat). The category $\mathsf{Mod}_{\Bbbk}$ is now an abelian monoidal category with exact tensor product. Dualizable object should be just the finitely-generated and projective ones. If $f:M\to N$ is a surjective morphism of $\Bbbk$-modules, then $\ker(f)$ is finitely-generated and projective, but:

Q1: what happens if $f$ is not surjective? Is $\ker(f)$ still finitely-generated and projective?

Q2: if $f$ is injective, what can we say about $\operatorname{coker}(f)$?

To answer to question 3, let $\mathscr{A}$ be an abelian (strict) monoidal category with exact tensor product and let $f:X\rightarrow Y$ be a morphism between (right) rigid objects in $\mathscr{A}$. 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 $\mathsf{coev}_{k}:\mathbb{I}\rightarrow \mathrm{coker}\left( f\right) ^{{\star }}\otimes \ker \left( f\right)$ as follows. Consider the composition $$\mathbb{I}\overset{\mathsf{coev}_{X}}{\longrightarrow }X{^{\star }}\otimes X\overset{c_{{\star }}\otimes X}{\longrightarrow }\mathrm{coker} \left( f^{{\star }}\right) \otimes X.$$ We have that \begin{align*} \left( \mathrm{coker}\left( f^{{\star }}\right) \otimes f\right) \circ \left( c_{{\star }}\otimes X\right) \circ \mathsf{coev}_{X} &=\left( c_{{\star }}\otimes Y\right) \circ \left( X^{{\star }}\otimes f\right) \circ \mathsf{coev}_{X} \\ &=\left( c_{{\star }}\otimes Y\right) \circ \left( f{^{\star }}\otimes Y\right) \circ \mathsf{coev}_{Y}=0 \end{align*} and hence $\left( c_{{\star }}\otimes X\right) \circ \mathsf{coev}_{X}$ factors through the kernel of $\mathrm{coker}\left( f{^{\star }}\right) \otimes f$, i.e. there exists a unique $\mathsf{coev}_{k}$ such that $$\left( c_{{\star }}\otimes X\right) \circ \mathsf{coev}_{X}=\left( \mathrm{ coker}\left( f^{{\star }}\right) \otimes k\right) \circ \mathsf{coev}_{k}.$$ Secondly, we define $\mathsf{ev}_{k}:\ker \left( f\right) \otimes \mathrm{ coker}\left( f^{{\star }}\right) \rightarrow \mathbb{I}$ as follows. Consider the composition $$\ker \left( f\right) \otimes X^{{\star }}\overset{k\otimes X^{{\star }}}{ \longrightarrow }X\otimes X^{{\star }}\overset{\mathsf{ev}_{X}}{ \longrightarrow }\mathbb{I}.$$ 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{coev}_{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{coev}_{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{coev}_{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{coev} _{X}\right) \\ &=\left( \mathsf{ev}_{X}\otimes X\right) \circ \left( X\otimes \mathsf{coev} _{X}\right) \circ k=k, \end{align*} from which we deduce (since $k$ is mono) that $\left( \mathsf{ev}_{k}\otimes \ker \left( f\right) \right) \circ \left( \ker \left( f\right) \otimes \mathsf{coev}_{k}\right) =\mathrm{id}_{\ker \left( f\right) }$, and \begin{align*} \left( \mathrm{coker}\left( f^{{\star }}\right) \otimes \mathsf{ev} _{k}\right) & \circ \left( \mathsf{coev}_{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{coev}_{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{coev}_{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{coev}_{X}\otimes X^{{\star }}\right) \\ &=c_{{\star }}\circ \left( X^\star\otimes \mathsf{ev}_{X}\right) \circ \left( \mathsf{coev}_{X}\otimes X^{{\star }}\right) =c_{{\star }}, \end{align*} from which we deduce that $\left( \mathrm{coker}\left( f{^{\star }}\right) \otimes \mathsf{ev}_{k}\right) \circ \left( \mathsf{coev}_{k}\otimes \mathrm{ coker}\left( f^{{\star }}\right) \right) =\mathrm{id}_{\mathrm{coker}\left( f^{{\star }}\right) }$.