Exercise 1.2.23 https://arxiv.org/pdf/1612.09375.pdf
Let $G$ be a group, regarded as a one-object category all of whose maps are
isomorphisms. Then its opposite $G^{op}$
is also a one-object category all of
whose maps are isomorphisms, and can therefore be regarded as a group
too. What is $G^{op}$, in purely group-theoretic terms? Prove that $G$ is isomorphic to $G^{op}$.
[Feel free to skip this part — I think it's okay, — but if you read it and find some mistakes, let me know.] Solution of part 1
("description in purely group-theoretic terms"):
Let $\mathcal G$ be the category corresponding to $G$. As a set, $G^{op}$ is still the set of morphisms in $\mathcal G^{op}$. So if the composition in $G$ is defined by $gh=g\circ h$, then that in $G^{op}$ is given by $gh=h^{op}\circ_{op} g^{op}$. The unit $1_G$ in $G$ is the identity morphism on the unique object of $\mathcal G$, so the unit in $G^{op}$ is $1_G^{op}$ (that this is an identity arrow in $\mathcal G^{op}$ can be proved by using $r^{op}\circ_{op} s^{op}=(s\circ r)^{op}$). Finally, an inverse for $g^{op}\in G^{op}$ is given as follows: the arrow $g^{op}$ in the category $\mathcal G^{op}$ is an isomorphism; let $g^{-op}$ be its category-theoretical inverse. The very same $g^{-op}$ is the group-theoretic inverse of
$g^{op}\in G^{op}$.
Regarding an isomorphism: We need to define a functor $F: \mathcal G\to\mathcal G^{op}$ and a functor $H: \mathcal G^{op}\to \mathcal G$ such that $F\circ H$ is the identity functor on $\mathcal G^{op}$ and $H\circ F$ is the identity functor on $\mathcal G$. It's clear how $F$ and $H$ behave on objects (on the unique object of each category). So we only need to say what they do to morphisms. If $f$ is a morphism in $\mathcal G$, let $F(f)=f^{op}$. Similarly, if $f^{op}$ is a morphism in $\mathcal G^{op}$, let $H(f^{op})=f$.
Let's try to see why $F$ is a functor (the proof that $H$ is one should be similar). We need to show that $F(f\circ g)=F(f)\circ_{op} F(g)$ and $F(id)=id$ whenever $f\circ g$ makes sense. That $F(id)=id$ is clear (we need to verify that $F(id)$, which is by definition $id^{op}$, is the identity arrow in $\mathcal C^{op}$; this follows from the equality $r^{op}\circ_{op} s^{op}=(s\circ r)^{op}$ and the fact that $id$ is a category-theoretical identity in $\mathcal G$). Let's postpone the verification of functoriality.
To see that $F\circ H$ is the identity functor, it suffices to show that it is the identity function on the set/class of arrows of $\mathcal G^{op}$. This is clear: an arrow $f^{op}$ is mapped to $f$ via $H$ and then back to $f^{op}$ via $F$.
Let's return to functoriality, the only thing that remains to show. We have $F(f\circ g)=(f\circ g)^{op}=g^{op}\circ_{op} f^{op}=F(g)\circ_{op} F(f)$… That's not what we need, right? Are my definitions of $F$ and $H$ wrong? The rest seems to be working fine. (Note: in the source of the exercise, "functor"="covariant functor".)
Also, does the proof use that all arrows in the category are isomorphisms? I don't see where it's used.
Best Answer
In group theoretical terms, $G^{op}$ is the same as $G$ but with multiplication $x\cdot_{op} y=yx$. As you see, not even at this level the identity is a homomorphism, since $f(x)\cdot_{op}f(y)=x\cdot_{op} y=yx\neq f(xy)=xy$.
An isomorphism at a group theoretical level that you can translate into the categorical language is $f:x\mapsto x^{-1}$. This way, $f(xy)=(xy)^{-1}=y^{-1}x^{-1}=x^{-1}\cdot_{op}y^{-1}=f(x)\cdot_{op}f(y)$ as you wanted.
In the categorical setting, this is just sending each arrow to its inverse arrow in the oposite category, i.e. $F(f)=(f^{op})^{-1}$. This way you have
$F(f\circ g)=(g^{op}\circ_{op} f^{op})^{-1}=(f^{op})^{-1}\circ_{op}(g^{op})^{-1}=F(f)\circ_{op} F(g).$
By the way, this does require that every morphism has an inverse. As an bonus exercise, try to show that every antiautomorphism $G\to G$ gives rise to an isomorphism $G\to G^{op}$.