# Categories which are dually equivalent to themselves.

I'm new to category theory and I'm struggling to get an intuition for very basic concepts. Assuming the notation that we characterize dual equivalence for categories $A$ and $B$ by an equivalence functor(<=> full, faithful, isomorphism-dense/essentially surjective) between $A^{\text{op}}$ and $B$.

General questions:

1.a. Is there some deeper meaning to the property of a category to be dually equivalent to itself? I.e. has the sub-conglomerate of these categories in $Cat$ important properties?

1.b. In what respect does dual equivalence of a category to itself generalize self-duality?

Examples:

2.a. Are there some instructive examples for categories which are dually equivalent to themselves?

Only the category of finite dimensional vector spaces and categories which are groupoids(even dually isomorphic to themselves) come to my mind (e.g. category of relations $Rel$, sets as objects, relations as morphisms).

Counterexamples:

So far it seems to me, that the categories of sets $Set$(and of finite sets $FinSet$), topological spaces $Top$, magmas $Mag$, Semigroups $Sem$, sets structured with relations $Rel_n Set$ (sets with $n$-relations as objects and maps perserving these as morphisms) are not dually equivalent to themselves since they have initial/terminal objects which aren't zero objects.

3.a. Why is the category of $\mathbb{k}$-vector spaces $Vec_{\mathbb{k}}$ not dually equivalent to itself?

I would argue: Assume an equivalence functor $F: Vec_{\mathbb{k}} \rightarrow Vec_{\mathbb{k}}^{op}$, $\aleph_0$-dim. $V_0$ and $\aleph_1$-dim. $V_1$ as $Vec_{\mathbb{k}}$-objects. To be full/faithful $F$ has to map infinite-dim. vectorspaces within their isomorphism-classes, i.e. $V_i\cong F(V_i)$. But since equivalences preserve/reflect mono/epimorphisms I get a contradiction by duality: $\emptyset=epi_{Vec_{\mathbb{k}}}(V_0,V_1)\cong epi_{Vec_{\mathbb{k}}^{op}}(F(V_0),F(V_1))=mono_{Vec_{\mathbb{k}}}(F(V_1),F(V_0))\neq \emptyset$.

I would like to argue similarly for the category of groups $Grp$, but I have no idea about the general cardinality of hom-sets in $Grp$, so...

3.b. Why is the category of groups $Grp$ not dually equivalent to itself?

3.c. What about the categories of monoids $Mon$?

3.d. Are there more instructive counterexamples?

• Two errors. If $\aleph_0\leqslant \dim(U)<\dim(V)$, it does not follow that $|\mathrm{end}(U)|<|\mathrm{end}(V)|$. If $|U|<|V|$ and $|F(U)|<|F(V)|$, then $\mathrm{epi}(U,V)=\emptyset=\mathrm{mono}(F(V),F(U))$, so no contradiction. May 24, 2021 at 23:45
• Also note that, for a category to be self dual, it is not necessary that terminal and initial objects coincide. Consider, for example, the category with two objects and a single arrow between them. May 25, 2021 at 2:49
• are you making a difference between a category being "self-dual" and "dually equivalent to itself"? May 25, 2021 at 2:54

If the category of groups was equivalent to its opposite, both would be semiabelian (since the definition of semiabelian in non-evil); but then (click, §1.4) $\bf Grp$ would be abelian, and it's not.

You can see it's a general argument, so also $\bf Mon$ isn't self-dual, and more generally several semiabelian varieties of algebras which are not abelian categories (there are many! See Janelidze and others' book).

You ask for nontrivial instructing examples: I vaguely remember the following result which relates self-duality to a strong bound on cardinalities of hom-sets.

Let $\cal C$ be a locally small cartesian closed with an initial object $\varnothing$. Define the "dual" of an object $X\in\cal C$ to be $[X, \varnothing]$: this is a functor ${\lnot}\;\colon {\cal C}^\text{op}\to \cal C$. If this functor is an equivalence, then every $\hom(X,Y)$ is either empty or a sigleton.

The solution for the category $$\mathbf{Vec}_k$$, proposed by the author of the question, is faulty, as I pointed out in a comment.

Here is a solution, which correctly derives a discrepancy between two cardinal numbers that should be equal. The idea is that an equivalence $$S\colon\mathbf{Vec}_k\to\mathbf{Vec}_k^\mathrm{op}$$ maps the copower $$k^{(I)}=\coprod_{i\in I}k^1=\bigoplus_{i\in I}k^1$$ in $$\mathbf{Vec}_k$$ of the one-dimensional vector space $$k^1$$ (the field $$k$$ regarded as a vector space over itself) to the copower $$\coprod_{i\in I}V$$ in $$\mathbf{Vec}_k^\mathrm{op}$$ of the vector space $$V=S(k^1)$$, but this copower is in $$\mathbf{Vec}_k$$ the power $$\prod_{i\in I}V\!=V^I\!$$. Assuming that $$I$$ is infinite and that $$|I|\geq|k|$$, we shall prove that $$\bigl|\mathrm{end}\bigl(V^I\bigr)\bigr| > \bigl|\mathrm{end}\bigl(k^{(I)}\bigr)\bigr|$$ (the first set of endomorphisms is in $$\mathbf{Vec}_k$$, while the second set of endomorphisms is in $$\mathbf{Vec}_k^\mathrm{op}$$ which, however, coincides with the set of endomorphisms in $$\mathbf{Vec}_k$$), a contradiction that proves that $$S$$ cannot exist.

We shall need the following result: if $$\mathfrak{h}$$ is an infinite cardinal, then $$\mathfrak{h}^\mathfrak{h}=2^\mathfrak{h}$$. This follows from $$\begin{equation*} 2^\mathfrak{h} \leqslant \mathfrak{h}^\mathfrak{h} \leqslant (2^\mathfrak{h})^\mathfrak{h} = 2^{\mathfrak{h}\cdot\mathfrak{h}} = 2^\mathfrak{h}~. \end{equation*}$$ Also recall that if $$F$$ is a field and $$\mathfrak{d}$$ is an infinite cardinal, then $$\begin{equation*} \dim(F^\mathfrak{d}) = |F^\mathfrak{d}| = |F|^\mathfrak{d}~. \end{equation*}$$

Now, the proof. First note that the vector space $$V$$ is not trivial, since $$S$$ maps $$\mathrm{end}(k^1)\cong k^1$$ bijectively onto $$\mathrm{end}(V)$$. Next, we have the bijections (with all end-sets and hom-sets in $$\mathbf{Vec}_k$$) \begin{align*} \mathrm{end}\bigl(k^{(I)}\bigr) &\cong \hom\bigl(k^1,\,k^{(I)}\bigr)^I~, \\[1ex] \mathrm{end}\bigl(V^I\bigr) &\cong \hom\bigl(V^I\!,V\bigr)^I~. \end{align*} Since $$\hom\bigl(k^1,\,k^{(I)}\bigr) \cong k^{(I)}$$ and $$\bigl|k^{(I)}\bigr| = \max(|k|,|I|) = |I|$$, we can calculate: $$\begin{equation*} \bigl|\mathrm{end}\bigl(k^{(I)}\bigr)\bigr| = |I|^{|I|} = 2^{|I|}~. \end{equation*}$$ The grand finale: $$\begin{gather*} \bigl|\mathrm{end}\bigl(V^I\bigr)\bigr| \:=\: \Bigl(|V|^{\dim(V^I)}\Bigr)^{|I|} \geqslant \Bigl(|k|^{\dim(k^I)}\Bigr)^{|I|} =\: |k|^{|k|^{|I|}\cdot|I|} \\[1ex] \geqslant\: 2^{2^{|I|}\cdot|I|} = 2^{2^{|I|}} >\: 2^{|I|} \:=\: \bigl|\mathrm{end}\bigl(k^{(I)}\bigr)\bigr|~. \end{gather*}$$ Done.