2
$\begingroup$

Am I correct in thinking that the category pGrp, whose objects are pairs $(G,g)$ where $G$ is a group, $g \in G$ and

$$\hom\left((G,g),(H,h) \right) = \{ \varphi: G \rightarrow H \hspace{1mm} \big\vert \hspace{1mm} \varphi(g) = h \},$$

is neither isomorphic to nor equivalent to the regular category Grp?

The counterexample I was thinking of was that $\hom\left((\mathbb{Z},2), (\mathbb{Z},3)\right) = \{ \} $, but that, under the forgetful functor $F$ which essentially forgets the points and leaves the groups and homomorphisms intact, the corresponding hom-set would be $\hom \left( \mathbb{Z},\mathbb{Z} \right) $, an infinite set (since we can have $\varphi(x) = nx $ for any $ n \in \mathbb{Z}$), so $F$ would not be full.

I suppose technically this wouldn't constitute a proof (it would only be a proof that $F$ is not an equivalence), since there are other functors from pGrp to Grp, but is my line of thinking generally correct?

Thanks!

$\endgroup$
3
  • $\begingroup$ I don't know enough category theory to tell you whether or not your line of thinking is correct. However, I think for the category $Grp$ the only initial/terminal/null/zero object is the trivial group, whereas for pointed groups we could have that 2-element groups are also terminal objects (if we select the non-identity object as the base point). en.wikipedia.org/wiki/Initial_and_terminal_objects From that it should follow (I imagine) that the two categories cannot be isomorphic. $\endgroup$ Commented Jun 18, 2016 at 22:42
  • 3
    $\begingroup$ @William If they exist, terminal objects in a category are unique (up to isomorphismus). Here the terminal object is $(\{e\},e)$. $\endgroup$ Commented Jun 18, 2016 at 23:32
  • $\begingroup$ @ChristianSievers Oh yeah that is a good point, I did learn that terminal objects are "unique up to unique isomorphism". Good thing I didn't post this as an answer then. $\endgroup$ Commented Jun 18, 2016 at 23:37

2 Answers 2

4
$\begingroup$

You are correct that they are not isomorphic nor equivalent, and your basic idea is along the right lines; you just may want to try and argue with some of the "special" objects in $\mathbf{Grp}$, like the zero object, to ensure that there can be no isomorphism nor equivalence. Consider that in $\mathbf{Grp}$ the group $1$ is both initial and terminal. However, in $\mathbf{Grp}_\ast$, the category of pointed groups, we may want to consider maps of the form $$\phi:(1,1) \to (\lbrace e, g \rbrace, g),$$ where $(1,1)$ is the trivial group with the obvious point identified and $\lbrace e, g \rbrace$ is a group isomorphic to $\mathbb{Z}/2\mathbb{Z}$ with $e$ the identity element. In order for $\phi$ to be a homomorphism in $\mathbf{Grp}_\ast$, $\phi$ must first be a homomorphism of groups (in the usual sense) and secondly must send the identity element of $1$, to the non-identity element of $\lbrace e, g \rbrace$. But then we can easily verify that the assignment $\phi(e) := g$ is not a homomorphism of groups and hence $$ \mathbf{Grp}_\ast((1,1),(\lbrace e, g \rbrace, g)) = \emptyset;$$ we conclude that the trivial group is not initial in $\mathbf{Grp}_\ast$. This shows that in particular the trivial group cannot be a zero object in $\mathbf{Grp}_\ast$ and hence that there can be no equivalence nor isomorphism between the categories.

$\endgroup$
3
  • 1
    $\begingroup$ One might suspect that there may be a different zero object. However, $(1,1)$ is obviously a terminal object in $\mathbf{Grp}_*$, but it is not initial, so the category has no zero object. $\endgroup$
    – egreg
    Commented Jun 18, 2016 at 23:29
  • $\begingroup$ @egreg If there was a zero object, no hom set could be empty. $\endgroup$ Commented Jun 18, 2016 at 23:35
  • $\begingroup$ @ChristianSievers Right, but I was commenting on Geoff's argument $\endgroup$
    – egreg
    Commented Jun 19, 2016 at 8:48
4
$\begingroup$

Geoff's answer is correct, and you are mostly right, too.

However, we have You now agree that $$\hom((\mathbb{Z},2),(\mathbb{Z},3))=\emptyset,$$ and that alone shows that this category is different from the category of groups (because that is connected).

Let me note that I have never seen this category before. With the identity element, groups have a "natural" point. Usually we point categories like the one of sets, of metric or of topological spaces, thereby creating a zero object. In this case, we destroy the zero object. That's really interesting!


By the way, your category is isomorphic (sic!) to the comma category $$(\mathbb{Z}\downarrow\mathcal{Grp})$$


Also note that you deviate from standard terminology. Usually, when you have a category $\mathcal{C}$ with a terminal object $1$, a pointed object of that category is an object $A$ together with a morphism $1\to A$. Morphisms to another such pointed object $1\to A'$ are morphisms $A\to A'$ that make the obvious triangle commutative. That is the category $(1\downarrow\mathcal{C})$. It has a zero object, and if $\mathcal{C}$ alreay had a zero object, both categories are isomorphic.

Your alternative generalization of pointed sets seems to work like this: let $\mathcal{C}$ be a concrete category (a category of structured sets), that is a category with a faithful functor $U:\mathcal{C}\to\mathcal{SET}$. A pointed object is an object $A$ of that category together with an element $a\in U(A)$. A morphism $(A,a)\to(A',a')$ is a morphism $A\to A'$ that satisfies $U(f)(a)=a'$. We have seen that this construction has less nice properties. On the other hand, one may argue that only this construction is of interest if $\mathcal{C}$ already has a zero object. In any case, if $\mathcal{C}$ has a free object $F_1$ over a singleton set, then we get a category that is isomorphic to $(F_1\downarrow\mathcal{C})$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .