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Could someone explain this to me in basic terms and give examples explaining why? Also, am I correct in that the initial object in a group is also the terminal object b/c all morphisms are reversible?

rus365
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  • What is your definition of a initial/terminal object in a group? – azif00 Dec 16 '20 at 04:42
  • initial object: object with a unique morphism from it to all other objects; terminal object: object with unique morphisms from all other objects to itself – rus365 Dec 16 '20 at 04:44
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    do you perhaps mean a groupoid, rather than a group? – Atticus Stonestrom Dec 16 '20 at 04:45
  • hm, I do have the same question for groupoids (which are like groups but have more than one object?) but thought I would think through the case for groups first – rus365 Dec 16 '20 at 04:46
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    ah, so the answer to your question is no, unless the group is trivial. indeed, if $G$ is a non-trivial group, then the unique object in the groupoid associated to $G$ will have more than $1$ morphisms to itself – indeed, it will have one for every element of the group! in a category, an initial object must have a unique morphism to every other object; it cannot have more than one morphism to another object. (and likewise the dual statement holds for terminal objects) – Atticus Stonestrom Dec 16 '20 at 04:47
  • sorry, what is the unique element in the groupoid associated with G? er and how do I get from G to the groupoid associated with G? – rus365 Dec 16 '20 at 04:49
  • well, the usual definition of a group is a set equipped with a binary operation satisfying some properties, not a category. to each group there is a natural way to associate a corresponding groupoid, which has a single object and one automorphism of that object for every element of the group, so one could technically define a "group" as just being a groupoid with a single object, but I don't think this is often done – Atticus Stonestrom Dec 16 '20 at 04:51
  • however, if the definition of a "group" that you're going with is just a groupoid with a single object, then the argument above still works – if the group has any morphism other than the identity morphism, then it will have no initial or terminal objects – Atticus Stonestrom Dec 16 '20 at 04:55
  • ok, so if i had the group (N, +), then 0 would be the initial and terminal object? And wouldn't this group have more than one object? -- I think I'm misunderstanding what an object is :/ could you explain this example? – rus365 Dec 16 '20 at 05:04
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    $(\mathbb{N},+)$ isn't a group, but a monoid, and it also has no initial or terminal object for the same reason no non-trivial group has these things. – Malice Vidrine Dec 16 '20 at 05:06
  • as Malice points out, $(\mathbb{N},+)$ is not a group; probably you were thinking of $(\mathbb{Z},+)$. to answer your question it would be helpful if you told us what your definition of a group is; is it the definition given in the wikipedia article I linked to, or is it "a groupoid with one object"? because, in the former case, it doesn't make sense to speak of "initial" or "terminal" "objects" of the group – a set is not naturally a category – Atticus Stonestrom Dec 16 '20 at 05:09
  • Let's go with a groupoid with one object. And yes, i realized that (Z, +) is what I should have written. Thanks for answering so many basic questions – rus365 Dec 16 '20 at 05:11
  • no worries, happy to help :) ... okay, so, let's consider $(\mathbb{Z},+)$ as a groupoid with one object; call that object $\star$. then there is a morphism $f_n:\star\rightarrow\star$ for every $n\in \mathbb{Z}$; we define a composition law by $f_m\circ f_n=f_{m+n}$. it is straightforward to verify that this satisfies the associativity axioms required in a category, and that $f_0$ will be the identity morphism on $\star$ – Atticus Stonestrom Dec 16 '20 at 05:13
  • now, this groupoid has only a single object – namely, $\star$ – and so that object is the only candidate for being an initial or a terminal object. however, $\star$ has many distinct morphisms to itself (one for every element of $\mathbb{Z}$!); since an initial or terminal object must have only a single morphism to itself, this means that $\star$ cannot be initial or terminal, and hence our groupoid has no initial or terminal objects – Atticus Stonestrom Dec 16 '20 at 05:14
  • I see, so there are many morphisms from * to itself so it cannot be initial/terminal. But inside *, these morphisms are all from elements of the integers to each other. Soo... any groupoid with an object with more than one underlying element cannot have initial/terminal elements? – rus365 Dec 16 '20 at 05:18
  • so, in the groupoid we've defined above, it doesn't make sense to think of things "inside" $\star$; in arbitrary categories, the objects don't necessarily need to have "elements". the group is just the collection of automorphisms of $\star$; it's not "elements" living inside $\star$, which is really just a dummy object containing no elements. this link discusses the notion that's confusing you; hopefully it helps a bit – Atticus Stonestrom Dec 16 '20 at 05:21
  • the object as a 'thing which has as its group of symmetries the group ' -- that makes sense now, thanks – rus365 Dec 16 '20 at 05:23
  • my pleasure :) ... so, just to sum up, a groupoid with one object $\star$ will have an initial or terminal object if and only if the identity morphism is the only automorphism of $\star$ – Atticus Stonestrom Dec 16 '20 at 05:24
  • ok, got it!! but what if the groupoid has multiple objects? – rus365 Dec 16 '20 at 05:27
  • sounds good, will ponder this. Thanks! – rus365 Dec 16 '20 at 05:52
  • happy to help! :) – Atticus Stonestrom Dec 16 '20 at 05:52

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If the group is seen as a category on one object $*$ (and the group elements are considered as the arrows $*\to *$), then, as there's no more object, if it contains an initial object, it must be $*$ but that would imply that there's only one arrow $*\to *$, i.e. the group is trivial.

If a groupoid (i.e. a category where every arrow is invertible) has an initial object $0$ then it's in particular connected, there's exactly one arrow $0_x:0\to x$ for any object $x$, and thus, for any objects $x,y$ we get $0_y\circ 0_x^{-1}$ as the only arrow $x\to y$.

If $f:x\to y$ then $f\circ 0_x:0\to y$ so by $0$ being initial, we get $f\circ 0_x=0_y$. Then use that $0_x$ is invertible.

So, a groupoid has initial object iff there's exactly one arrow between any two objects iff every object is initial iff every object is terminal iff it has terminal object iff it is equivalent to the terminal category (the discrete category on one object).

On the other hand, the category of groups and homomorphisms does have a zero object (which is initial and terminal at once), namely the trivial group $\{1\}$.
Assigning $1\mapsto 1_G$ is the unique homomorphism $\{1\}\to G$ to any group, and the constant $1$ map is the unique homomorphism $G\to\{1\}$ from any group.

Berci
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