2
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ @Javi So is it unnecessary to describe inverses and the unit? Technically, the unit in $G^{op}$ is different from that in $G$, as far as I understand (it corresponds to the reversed arrow that in turn corresponds to the unit in $G$). $\endgroup$
    – user634426
    Jun 20 '19 at 23:13
  • $\begingroup$ In group theoretic terms, the identity is the same. I'll try to write an answer with a natural isomorphism. $\endgroup$
    – Javi
    Jun 20 '19 at 23:15
  • $\begingroup$ The isomorphism from $G$ to $G^{op}$ has to send $x$ to $x^{-1}$. It's probably easier to understand this by doing the group theory first and then translating into category theory (if you feel you need to do that). $\endgroup$
    – Rob Arthan
    Jun 20 '19 at 23:16
  • $\begingroup$ @Javi Then I think I don't quite understand what is meant by "group-theoretic". Also, natural transformations appear later than this exercise in the source, so I would prefer to avoid using them. $\endgroup$
    – user634426
    Jun 20 '19 at 23:18
  • $\begingroup$ Group theoretic means that you have a set with a binary operation. The identity is an element which acts trivially under this operation, and it is the same in both groups (the underlying set is the same, so it is not just the image under some homomorphism). $\endgroup$
    – Javi
    Jun 20 '19 at 23:27
2
$\begingroup$

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 $G\to G^{op}$.

$\endgroup$
6
  • $\begingroup$ Then should $H$ be similarly defined as $f^{op}\mapsto f^{-1}$? If so, I don't see why $F\circ H$ is the identity functor anymore. $f$ gets mapped to $(f^{op})^{-1}$ by $H$. This thing is supposed to be mapped to $f^{op}$ by $F$, why is this the case? $\endgroup$
    – user634426
    Jun 20 '19 at 23:51
  • $\begingroup$ Define $H(f^{op})=f^{-1}$. Then $F\circ H(f^{op})=F(f^{-1})=((f^{-1})^{op})^{-1}=f^{op}$. $\endgroup$
    – Javi
    Jun 21 '19 at 8:31
  • $\begingroup$ I also got $F\circ H(f^{op})=((f^{-1})^{op})^{-1}$, but how did you get the last equality? $\endgroup$
    – user634426
    Jun 21 '19 at 14:07
  • 1
    $\begingroup$ You can check directly that the inverse of the opposite arrow is the opposite of the inverse. If $g^{-1}\circ g=1$, then taking the opposite yields $g^{op}\circ_{op}(g^{-1})^{op}=1$, so $(g^{-1})^{op}=(g^{op})^{-1}$. $\endgroup$
    – Javi
    Jun 21 '19 at 16:06
  • $\begingroup$ Now I understand why $F(f\circ g)=F(f)\circ_{op} F(g)$ holds, but if we use the superscript $^{op}$ to indicate that the arrow being considered lives in $\mathcal G^{op}$, why don't we equip the $F(f)$ and $F(g)$ in the equality above with that superscript? They do live in $\mathcal G^{op}$. $\endgroup$
    – user634426
    Jun 21 '19 at 21:59

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.