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 $\operatorname{coker}(f)$?


It seems that asking helps in enlightening: at the end I found an answer at least for questions 1 and 3. The answer to question 1 is: yes, it was so trivial that I was just sinking in a glass of water.

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

The other one is analogous and hence the claim holds (in particular, also for modules over a von Neumann regular ring).

Since an answer to question 2 would just have been a step in proving the foregoing facts, it is not needed anymore. Nevertheless, since knowing an answer for that question as well could be interesting, I will maybe ask it separately and consider this question answered.

In case somebody will find a reference as in question 4, I am still interested. So, please, answer here or comment if you find it. Many thanks.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.