Notation: Underlining $\underline{G}$ denotes a category and $\underline{G}(x,y)$ the class of morphisms from $x$ to $y$.

On the Wiki page about groupoids, it is written (I write here my own more concise version):

Algebraic Definition: A partial function $f:X\to Y$ is a function $f:A\to Y$ for some $\emptyset\neq A\subseteq X$. A groupoid is a set $G$ with a unary operation $^{-1}:G\to G$ and a partial function $\ast:G\times G \to G$ that satisfy the following axiomatic properties for arbitrary $a,b,c\in G$:

  1. (associativity) If $a\ast b$ and $b\ast c$ are defined, then $(a\ast b)\ast c$ and $a\ast(b\ast c)$ are defined and equal. Conversely, if either of these last two expressions is defined, then so is the other, and again they are equal.
  2. (inverses) $a^{-1}\!\ast a$ and $a\ast a^{-1}$ are always defined.
  3. (identity) If $a\ast b$ is defined, then $a\ast b\ast b^{-1} \!= a$ and $a^{-1}\!\ast a\ast b = b$. (1. and 2. already show that these expressions are defined.)

In short:

  1. $(a\ast b)\ast c = a\ast(b\ast c)$;
  2. $\forall a \;\,\exists\, a^{-1}\!\ast a \;\,\exists\, a\ast a^{-1}$;
  3. $a\ast b\ast b^{-1} \!= a$ and $a^{-1}\!\ast a\ast b = b$.

From these axioms, two easy and convenient properties follow:

  1. $(a^{-1})^{-1} \!= a$;
  2. If $a\ast b$ is defined, then $(a\ast b)^{-1}\! = b^{-1}\!\ast a^{-1}$.

Category-Theoretic Definition: A groupoid $\underline{G}$ is a small category (i.e. its class of objects $\underline{G}$ and its class of morphisms $\mathrm{Hom}_{\underline{G}}$ are sets) in which every morphism is an isomorphism (i.e. is invertible). Thus a groupoid is equipped with the composition $\circ$ and invertion $^-$ of morphisms.

Comparing the Definitions: Both definitions are equivalent, as follows.

Given a category-theoretic groupoid $\underline{G}$, let $G$ be the set of all morphisms of $\underline{G}$. Then $\circ$ becomes a partially defined operation on $G$, and $^-$ is in fact defined everywhere. Therefore we take $\ast:=\circ$ and $^{-1}:=^-$ and obtain an algebraic groupoid.

Conversely, given an algebraic groupoid $G$, define the objects of a category $\underline{G}$ as $\{x\ast x^{-1};\:x\!\in\!G\}$ and define $\underline{G}(x,y)$ as $\{f\!\in\!G;\:y\ast f\ast x\text{ is defined}\}$. Then $^{-1}$ and $\ast$ break up into several functions on the various $\underline{G}(x,y)$, which may be called $^-$ and $\circ$, respectively. EDIT: the notation in this paragraph is confusing; see answers.

Question 1: What are the details of the last paragraph? Why are objects $x\ast x^{-1}$ and not just $x$? How do we compose two morphisms $f$ and $g$, when we know $z\ast g\ast y$ and $y\ast f\ast x$ exist? As $g\ast y\ast f$? Or just $g\ast f$? What are the identity morphisms?

Comment: I realise this is an elementary question, but it is surprisingly hard to find anything in the books about groupoids. Almost no Abstract Algebra book contains anything about groupoids. I'm dealing with this notion for the first time. The only book I located that deals with groupoids is Topology and Groupoids by Ronald Brown, but I can't seem to find the algebraic definition in it.

Question 2: What is a subgroupoid? A subcategory that is itself a groupoid? What is a groupoid homomorphism? An arbitrary functor between groupoids?

Question 3: (a too broad question, I know, but anything will be useful) Are there any alternate definitions of a groupoid? I'm aiming for one that is used most often throughout mathematics nowadays. How/in what fields are groupoids most often used? Algebraic topology? Are they given topological structure? What are some important theorems regarding groupoids?

  • 4
    $\begingroup$ You may want to look at P.J. Higgins's classic text Categories and Groupoids. It contains a lot of basic structure theory and constructions for groupoids at a very explicit level. There are quite a few references on how groupoids are used in practice. The first few chapters of May's Concise course on algebraic topology also contains a lot. Look also at Weinstein's Groupoids: Unifying Internal and External Symmetry. $\endgroup$ – t.b. Aug 25 '11 at 16:20
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    $\begingroup$ I think part of your confusion is that when we define the objects, we are using $x*x^{-1}$. That means that in $\underline{G}(x,y)$, $x$ is actually equal to $z*z^{-1}$ for some $z\in G$, and $y$ is equal to $w*w^{-1}$ for some $w\in G$. So $f\in\underline{G}(x,y)$ means that $w*w^{-1}*f*x*x^{-1}$ is defined. Then if $g\in\underline{G}(w*w^{-1},t*t^{-1})$, then $t*t^{-1}*g*w*w^{-1}$ is defined; the composition is given by their "product", which is $t*t^{-1}*(g*f)*x*x^{-1}$. This is part of the reason why the object is defined as $x*x^{-1}$. Also, $x*x^{-1}$ is the identity of $x*x^{-1}$. $\endgroup$ – Arturo Magidin Aug 25 '11 at 16:49
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    $\begingroup$ The definitive introduction to the study of groupoids and their applications to topology is Ronald Brown's TOPOLOGY AND GROUPOIDS, Leon. If you want to learn about groupoids,that's really the place to start by one of the founders of their study. What's fascinating about Brown's book and it's "cousins"-like the new treatise NONABELIAN ALGEBRAIC TOPOLOGY-is that it allows for a nearly complete development of algebraic topology entirely in terms of homotopy-and homology groups and homological algebra ARE NEVER MENTIONED! $\endgroup$ – Mathemagician1234 Aug 25 '11 at 17:04
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    $\begingroup$ @Leon: (cont) It's very bad form to use $x$ to mean two different things in the same formula. The presentation you give suffers because of this: when defining the objects of $\underline{G}$, we are using $x$ to denote an element of $G$; but when we are defining the arrow sets $\underline{G}(x,y)$, we are using $x$ to denote an object of $\underline{G}$. That's why I used a different letter. So, when we write $\underline{G}(x,y)$, it's bad form to say "this is $\underline{G}(x*x^{-1},y*y^{-1})$, because now $x$ is both an object of $\underline{G}$ and an element of $G$, and likewise $y$. $\endgroup$ – Arturo Magidin Aug 25 '11 at 17:09
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    $\begingroup$ You also might find this survey by Brown very useful for the backstory on groupoids: pages.bangor.ac.uk/~mas010/groupoidsurvey.pdf $\endgroup$ – Mathemagician1234 Aug 25 '11 at 17:14

Much has been said already but I'd like to suggest to try and be a bit more careful in the formalization. Stars and underbars are not really used anywhere in the literature I've seen so far and I find them distracting. Also, minimization of the axioms is not always the royal way to clarity.

I'm going to give a detailed categorical definition of groupoids and then indicate how to relate it to the algebraic one you ask about and see that they are equivalent. I'll then briefly address your second question and towards the end I'll try and give some glimpses towards an answer of your broad question 3. I'm making my life a bit more complicated than necessary in that I don't introduce Hom-sets (or only implicitly) but that's because groupoids per se are much more interesting when considered in categories other than $\mathbf{Sets}$ and in order to be able to do so, the following definition is more adequate (view it as a preparation for my attempt at answering your question 3).

So here we go in full detail:

A groupoid $G$ consists of two sets: a set of objects $G_0$ and a set of arrows $G_1$ together with five structure maps $(s,t,u,i,c)$

  1. source and target $s,t: G_1 \to G_0$ (domain and codomain of arrows).
  2. unit $u: G_0 \to G_1$ (the unit inclusion mapping an object to its identity morphism)
  3. composition of arrows $c: G_2 \to G_1$, defined on the fibered product (pullback) $$G_{2} = G_1 {}_s{\times}_{t}\ G_1 = \{(h,g) \in G_1 \times G_1\,:\,s(h)=t(g)\}$$ and written $c(h,g) = hg$. (The set $G_2$ is the set of pairs of composable arrows).
  4. inversion of arrows $i: G_1 \to G_1$ written $i(g) = g^{-1}$

subject to the following conditions

  1. compatibility of source and target with composition: for all $(h,g) \in G_2$ we have $s(hg) = s(g)$ and $t(hg) = t(h)$.
  2. associativity: for all $k,h,g \in G_1$ such that $s(k) = t(h)$ and $s(h)=t(g)$ we have $(kh)g = k(hg)$ (note that this makes sense in view of 1).
  3. compatibility of unit with source, target and composition:
    • For all $x \in G_0$ we have $su(x) = x = tu(x)$
    • For all $g \in G_1$ we have $[ut(g)]g = g = g[us(g)]$
  4. compatibilities of the inverse: For all $g \in G_1$ we have $s(g^{-1}) = t(g)$ and $t(g^{-1}) = s(g)$ as well as $g^{-1}g = us(g)$ and $gg^{-1}=ut(g)$.

All this can be expressed in purely diagrammatic terms and leads to the notion of a groupoid inside a category (and makes sense as soon as there are enough pullbacks). Also when you flesh out the definition of a category on Wikipedia you see that the Hom-sets are simply $G(x,y) = \{g \in G_1:s(g) = x\text{ and }t(g) = y\}$ and that the structure maps above are implicitly built into the formalism.

I'm of the opinion that only this is a reasonable (category-theoretic) definition of groupoids and everything else is just cheating one's way around precision. When saying that one has a (small) category the usual formulation is exactly what is obtained when dropping $i$ and the two points 4. above.

So, when people are saying that "a groupoid is a small category in which every morphism is an isomorphism" then they mean first that they have a category, i.e., two sets $G_0$ and $G_1$ and four structure maps $(s,t,u,c)$ satisfying points 1., 2., and 3. above and 4. is what one ends up with when one thinks about what it really means for a(ll) morphism(s) to be invertible (isomorphisms). So much for the categorical part of your question 1.

Now I should relate this to the definition of algebraic groupoids. From the above it should be clear that given a groupoid in the above sense, $\ast = c$ is a partial operation satisfying associativity, that "inverses are always defined" and that the "identity" axiom is satisfied.

Now we need to check that we can go back from the algebraic definition to the category theoretic definition, so we need to identify the sets $G_1$ and $G_0$ and the structure maps. Here are the details (bear with me and let me drop the star):

  • Put $G_1 = G$ and $G_0 = \{g \in G: g^{-1} = g\}$ (idea: objects are the same thing as identity morphisms and they are distinguished by being their own inverses) the unit $u:G_0 \to G_1$ is simply the inclusion and inversion $i: G_1 \to G_1$ is already given.
  • Source is $s(g) = g^{-1}g$ and target is $t(g) = gg^{-1}$ these are well-defined because "inverses are always defined" and are maps $s,t: G_1 \to G_0$ because of your "easy and convenient property 2.".
  • It remains to identify the domain of definition of multiplication with $G_2$ above (that's already in Arturo's answer). As we already spelled out what source and target are, we need to check that $hg$ is defined whenever $h^{-1}h = s(h) \stackrel{!}{=} t(g) = gg^{-1}$ and conversely.

    If $h^{-1}h = gg^{-1}$ then $h = h(h^{-1}h)$ (take $a = b = h$ in the identity axiom) hence $h = h(h^{-1}h) = h(gg^{-1}) = (hg)g^{-1}$. But we can compose the last expression with $g$ from the right and get $(hg)g^{-1}g = hg$, hence $s(h) = t(g)$ implies that $hg$ is defined by the associativity axiom.

    Conversely if $hg$ is defined then $hgg^{-1} = h$ hence composing this with $h^{-1}$ from the left we get $s(g) = gg^{-1} = h^{-1}h = t(h)$.

With all this preparation you should now have no trouble verifying the axioms 1. to 4. of the categorical definition.

Question 2 is rather easy to answer:

A morphism $\phi: G \to G'$ between groupoids is just a functor. In the spirit of the above definition let me spell out the details. I'm using primes to denote the structure maps of $G'$:

It consists of two maps $\phi_{1} : G_1 \to G_1^\prime$ and $\phi_{0}: G_{0} \to G_{0}^\prime$ satisfying the following compatibilities with the structure maps:

  1. source and target: $\phi_0 s = s'\phi_1$ and $\phi_0 t = t' \phi_1$
  2. unit: $\phi_1 u = u' \phi_0$
  3. composition: if $s(h) = t(g)$ then $\phi_1(hg) = \phi_1(h)\phi_1(g)$.

A more formal way to express the composition axiom 3. would be to say that $\phi_1$ induces a unique map $\phi_2 = \phi_1 {}_s{\times}_{t}\ \phi_1: G_2 \to G_{2}^{\prime}$ such that $(s',t') \circ \phi_2 = \phi_1 \circ (s,t)$ and the condition then becomes $c' \circ \phi_2 = \phi_1 \circ c$.

Compatibility with inverses is then automatic: $\phi_1 i = i'\phi_1$.

A subgroupoid is usually defined to be a groupoid $H$ together with a functor $\phi: H \to G$ such that $\phi_1$ and $\phi_0$ are injective (embeddings). This means that $\phi_1: H_1 \to \phi_1(H_1)$ is an isomorhpism onto its image and so is $\phi_0: H_0 \to \phi_{0}(H_0)$.

A homomorphism of algebraic groupoids is simply a map $\phi: G \to H$ that is compatible with composition: whenever $hg$ is defined then so is $\phi(h)\phi(g)$ and it is equal to $\phi(hg)$. The description of arrows, objects, source, target unit and composition then yield that a homomorphism of algebraic groupoids yields a functor and vice versa. In particular the categories of algebraic groupoids and homomorphisms and the category of groupoids-as-categories are equivalent.

Note: Starting with an algebraic groupoid and passing to the associated category and going back to the algebraic setting, one gets the algebraic groupoid back. On the other hand, starting with a categorical groupoid and passing to the associated algebraic groupoid and going back one gets an isomorphic categorical groupoid but not necessarily the same one because we didn't insist on $G_0$ being a subset of $G_1$ in the categorical definition. One could add that requirement and get an isomorphism of categories, but neither do I see a compelling reason to do so, nor is it usually done, as far as I can tell.

One should also mention that not only are there functors but there are also natural transformations, so, as the category of (small) categories is not only a category but a $2$-category, so is the category of groupoids. But entering more detail would mean to go too far in this already very long answer. Let me just mention that this allows one to consider groupoids up to equivalence of categories; there is the even weaker notion of Morita equivalence that is of considerable interest from a geometric point of view.

Here are three basic examples of groupoids:

  1. Groups.
  2. Equivalence relations.
  3. Group actions.

A group is a groupoid in which $G_{0}$ is reduced to a singleton. An equivalence relation is a groupoid for which there is at most one morphism between any two objects. Group actions don't have such a handy characterization but to every action of a group $G$ on a set $X$ one can associate the so-called action groupoid (or translation groupoid) $G \ltimes X$.

Many theorems on groupoids had earlier variants for groups, equivalence relations and action groupoids and were then generalized later on. This is one criticism that one often hears: there are few results that go in the other direction, at least not on a basic or immediate level — some people see groupoids as a way of formalizing analogies that were already self-evident before.

As I mentioned in a comment, groupoids on a set-theoretic level are already interesting and the book by Higgins, Categories and Groupoids is a nice introduction to that with e.g. a groupoid version Grushko's theorem and explaining the use in algebraic topology via the fundamental groupoid with the groupoid version of van Kampen as a highlight.

This leads me to my second recommendation: May's Concise course in algebraic topology shows the use of groupoids in a nice formulation of the theory of covering spaces, among other things like van Kampen, of course.

Since you are already aware of Ronald Brown's work, I'm not elaborating on that. However I must say that I think that characterizing Brown as "one of the founders of their study" as was done in one of the comments is a bit unfair considering the earlier and deep work e.g. by Ehresmann and Haefliger upon which Brown's first work on groupoids was built (I don't know Brown personally but from his homage to Ehresmann it seems that he is more than happy to acknowledge that). At some point I should also mention that groupoids were introduced by Heinrich Brandt in 1940.

Your third question

Question 3: (a too broad question, I know, but anything will be useful) Are there any alternate definitions of a groupoid? I'm aiming for one that is used most often throughout mathematics nowadays. How/in what fields are groupoids most often used? Algebraic topology? Are they given topological structure? What are some important theorems regarding groupoids?

is very broad so I'll give a broad answer, just dropping some key-words and giving some links. I should also add the Disclaimer that I'm far from being an expert on groupoids I just happen to have run into them many times already. I'm not aspiring to any sort of comprehensiveness and I'm just mentioning things that came to mind immediately. One further problem is that groupoids are mostly used at a fairly advanced level, so mentioning some scary words is almost unavoidable. Also, the interesting theorems on (or using) groupoids I'm aware of all seem to involve baggage way beyond the scope of this question and this answer.

Briefly: It seems fair to say that in every sufficiently algebraized field there are people considering and using groupoids.

  • I'm not aware of significantly different definitions than the two you gave (but see e.g. this PlanetMath article on Brandt groupoids).
  • The category-theoretic one is certainly by far more often used than the algebraic one.
  • There are many versions of structured groupoids (smooth, topological, measure-theoretic and by means of algebro-geometric methods), as I mention below.
  • I can't tell where groupoids are most often used.

A very important, popular and thus highly developed field of investigation is in differential geometry via Lie groupoids. Lie groupoids simultaneously encompass Lie groups and manifolds and on a more advanced level arise from foliations in guise of the holonomy groupoids. This is one of the reasons I spelled out the definition in such painstaking detail above: In order to define a Lie groupoid one needs to be a bit careful on what conditions to impose on $G_1$, $G_0$ and the structure maps. The most straightforward definition ($G_0$ and $G_1$ manifolds and all maps smooth) turns out not to be quite the right one and in order to spell the "good" definition out one needs to do some careful bookkeeping and there's no way around writing down all the structure maps.

Lie groupoids have associated Lie algebroids which are of considerable interest too, very much so in mathematical physics.

As a first step into this direction I recommend the surveys by Alan Weinstein, Groupoids: unifying internal and external symmetry and The geometry of momentum. For a more in-depth view I recommend the book by Weinstein-Canas da Silva, Geometric models of non-commutative algebras for a start. There are several surveys by Moerdijk and Mrčun that you can locate easily if you want to dig deeper.

In representation theory they arise e.g. in the Tannakian formalism (I couldn't find a good and digestible link on that).

In ergodic theory and representation theory of locally compact groups they are used in guise of Mackey's virtual groups, roughly a theory built on a measure-theoretic version of groupoids.

In a topological and operator theoretic setting much has been done, mostly initiated by Jean Renault. For instance, J.-L. Tu used groupoids heavily for proving some forms of the Baum-Connes conjecture.

Not to mention category theory itself and advanced algebraic topology (see e.g. the work by Georges Maltsiniotis and his school elaborating on Grothendieck's approach to stacks), but I'll stop here because I don't like indulging in name-dropping and I've done already too much of it. I didn't mention or link to work on stacks in algebraic geometry (e.g. by Laumon and Moret-Bailly as well as Lurie and Toën, simply because I'm not qualified).

I hope this gave you some impression that there is a lot around and hopefully this answer gives you some leads you can follow.

  • $\begingroup$ Dear Theo, I found the analogy between Lie groupoids and Artin stacks very helpful: a Lie groupoid can be viewed as a "differentiable stack," while an algebraic stack is a groupoid in the category of algebraic spaces (where the structure maps are smooth). (This is a very detailed answer and I am simply pointing out a further analogy.) $\endgroup$ – Akhil Mathew Aug 28 '11 at 5:18
  • $\begingroup$ (By which I meant that any differentiable/algebraic stack has a presentation as a Lie/algebraic-space groupoid, not that the two categories are the same, of course.) $\endgroup$ – Akhil Mathew Aug 28 '11 at 5:20
  • $\begingroup$ @Theo: thank you very much for an outstanding answer. Sorry for the late reply, but I had to study other things in between, and it wasn't until now that I found the time to thoroughly and properly read your answer. Funny, it just so happens that Janez Mrčun (mentioned above) is a professor at my faculty :). Maybe I'll ask him, when the time is right. Thank you again, much appreciated. $\endgroup$ – Leo Sep 6 '11 at 22:51
  • $\begingroup$ @Leon: Thank you for the kind words and you're most welcome! That's great news that Mrčun is at your university! I met Ieke Moerdijk a few times and I'd bet that Janez Mrčun is similarly approachable. Don't be shy, talk to him soon, maybe he can point you to stuff that would lead you further. It would be great if you found some people with whom you can talk and I hope that this would help overcoming the situation you described in your question about healthy lifestyle. In any case, I wish you all the best and good luck in your pursuit of groupoids, if you should choose to dig deeper! $\endgroup$ – t.b. Sep 6 '11 at 23:15

Question 1. The reason you want the objects to be $x*x^{-1}$ and not just $x$ is so that the rules actually work out properly. For example, we clearly want $x*x^{-1}$ to be the identity map on the object corresponding to $x$. If we called the object "$x$", then $x*x^{-1}\in\underline{G}(x,x)$ if and only if $x*(x*x^{-1})*x$ is defined; but this requires $x*x$ to be defined, and this is not a given. However, $x*x^{-1}\in\underline{G}(x*x^{-1},x*x^{-1})$ if and only if $(x*x^{-1})*(x*x^{-1})*(x*x^{-1})$ is defined; and it is easy to verify it is defined, since $x*x^{-1}$, $(x*x^{-1})*x$, $((x*x^{-1})*x)*x^{-1}$, etc., are all defined.

This also allows you to compose two morphisms. If $f\in\underline{G}(x*x^{-1},y*y^{-1})$ and $g\in\underline{G}(y*y^{-1},z*z^{-1})$, then $y*y^{-1}*f*x*x^{-1}$ and $z*z^{-1}*g*y*y*^{-1}$ are both defined. Since $z*z^{-1}*g*y*y^{-1}*(y*y^{-1})$ and $(y*y^{-1})*f*x*x^{-1}$ are both defined, so is $$z*z^{-1}*g*y*y^{-1}*y*y^{-1}*f*x*x^{-1} = z*z^{-1}*g*f*x*x^{-1}$$ so $g*f$ makes sense and is in $\underline{G}(x*x^{-1},z*z^{-1})$, as desired.

This also shows you the identity morphisms are indeed given by $x*x^{-1}$: if $f\in \underline{G}(x*x^{-1},y*y^{-1})$, then $f*(x*x^{-1})$ makes sense and is equal to $f$, and likewise with $(y*y^{-1})*f = f$.

  • $\begingroup$ Ah yes, this makes a lot more sense. I'll edit wiki... $\endgroup$ – Leo Aug 25 '11 at 17:10

[Edit] This answer has been simplified and streamlined in reaction to an elementary observation by Theo Buehler in the comments, with regards to the nature of the construction.

In answer to your questions:

  1. (a) There is a convention used by some of identifying objects in a category with their own identity morphisms; the maps which take that object as the domain or codomain are those for which composition with the identity are well-defined. We can take advantage of this convention when dealing with an algebraic groupoid: for each $x$, the morphism $x \ast x^{-1}$ would be the map which satisfies the axioms of the identity operator with respect to (algebraic) groupoid elements that we can compose with $x$ on the left or right. So, needing to construct the category from the algebraic groupoid, we may simply define $x \ast x^{-1}$ to be the identity morphism for the object (and identity morphism) which is the domain/codomain of the maps composable with $x$.

    (b) It seems as though the composition of $f: (x \ast x^{-1}) \to (y \ast y^{-1})$ and $g: (y \ast y^{-1}) \to (z \ast z^{-1})$ would be $[g \ast f]: (x \ast x^{-1}) \to (z \ast z^{-1})$. The morphism condition should actually be read that $f \in \underline{G}(x \ast x^{-1}, y \ast y^{-1})$ if and only if $(y \ast y^{-1} \ast f \ast x \ast x^{-1})$ is well defined in the algebraic groupoid: then if $f \in \underline{G}(x \ast x^{-1}, y \ast y^{-1})$ and $g \in \underline{G}(y \ast y^{-1}, z \ast z^{-1})$ then the map

                    $(z \ast z^{-1} \ast g \ast y \ast y^{-1}) \ast (y \ast y^{-1} \ast f \ast x \ast x^{1}) \;=\; (z \ast z^{-1}) \ast g \ast f \ast (x \ast x^{-1})$

    is well defined, so that $g \ast f \in \underline{G}(x \ast x^{-1}, z \ast z^{-1})$.

  2. (a) A subgroupoid in the algebraic sense would just be a subset which is also a groupoid with the same composition operation. The category arising from this will also be a subcategory of the categorical groupoid. (A subcategory is essentially just a selection of some of the morphisms of the larger category, which contains the necessary objects, identity morphisms, and compositions to qualify as a category; given an algebraic subgroupoid, the very elements of the subgroupoid will ensure the presence of the necessary objects/identity-morphisms, and the composites in the algebraic subgroupoid ensures the presence of compositions in the subcategory.) Conversely, regarding the set of morphisms of a groupoidal category $C$ as an algebraic groupoid $A$, the morphisms of a subcategory $C'$ of the category $C$ will also be a groupoid $A'$ which is a subset of the algebraic groupoid $A$.

    (b) A morphism of subgroupoids, algebraically, is just a function between the groupoids which preserves composition. In order to express preservation of composition, we would need to speak categorically of a functor, as you correctly surmised; and the mapping does indeed define a functor, as it will map the domains and codomains (which we may identify with identity morphisms) appropriately. Conversely, any functor between categorical groupoids will preserve composition by construction, and thereby define a homomorphism of the groupoids defined by the morphism sets.

  • $\begingroup$ Aha, so the inverse of $f$ is $f^-:=f^{-1}$ and the identity morphisms are $1_{x\ast x^{-1}}=x\ast x^{-1}$. Would you perhaps be willing to address at least the second question, so I can accept your answer? $\endgroup$ – Leo Aug 25 '11 at 17:25
  • $\begingroup$ @Leon: There's no hurry in accepting an answer now (some people might want to add something when the initial flurry has calmed down). You received only a tiny little glimpse of what is possible towards an answer of your question. Also, concerning the editing of Wikipedia you announced in a comment to Arturo's answer may I suggest that you wait a little until the very basics on groupoids have sunk in? $\endgroup$ – t.b. Aug 25 '11 at 17:45
  • $\begingroup$ @Theo: Yeah, I always feel the pressure to accept an answer when one is given, thinking that people will find me too picky and ungrateful. As for Wiki, I only had in mind a minor edit, replacing the necessary $x$-es with $x\ast x^{-1}$-es in that paragraph alone. Otherwise, that's gonna confuse a lot more people. Anyway, thanks for advice :) $\endgroup$ – Leo Aug 25 '11 at 17:59
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    $\begingroup$ @Leon: Ah, okay. Then I trust your judgment. Concerning accepting an answer you may want to look here I would say this is one of those cases where a longer time may be indicated (no offense, Niel; I'm not saying your answer isn't good) $\endgroup$ – t.b. Aug 25 '11 at 18:06
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    $\begingroup$ @Leon: each set of morphisms is identified with $G$. So let me write $(x,g,y)$ for the morphism from $y$ to $x$ represented by $g \in G$. Then we can compose $(x,g,y)(y,h,z) = (x,gh,z)$. The identity elements are $(x,e,x)$, $(y,e,y)$, etc. The inverse of $(x,g,y)$ is then $(y,g^{-1},x)$ and so on. Do you get the picture? You store start and endpoint in a triple and you can only compose if start and endpoint "match". $\endgroup$ – t.b. Aug 26 '11 at 0:34

For your first question, sometimes, the best way to understand an equivalence like this is to try and do it yourself; consider the exercise:

  1. Suppose you are given an algebraic groupoid $H$, and are told that it was constructed from the set of morphisms of a category-theoretic groupoid $G$. (By the construction you gave in your question) From $H$, recover $G$ (up to isomorphism).

  2. Generalize the construction in exercise 1 to any algebraic groupoid.

For exercise 1, the set Morphisms($G$) and the composition law are obvious. The only challenge is recovering the set of objects, and figuring out which arrows belong in which hom-sets $G(x,y)$. There are lots of ways you might go about identifying objects; e.g. you might decide to specify that the objects are

Objects($G$) = { Source($x$) | $x \in H$ }

Of course, taking the source of an element of $H$ is cheating; instead, we should identify an object as an equivalence class [x], where [x] = [y] iff Source($x$) = Source($y$); we have some hope of finding an algebraic equation in $H$ that is equivalent to this equation.

We could choose other ways of identifying the objects. As others have pointed out, equating objects with identity morphisms -- morphisms $i$ with the property that $f\circ i=f$ and $i\circ g=g$ whenever the products are defined -- is a standard trick. In fact, one definition of category effectively defines Source($x$) to be an identity morphism such that $x \circ \text{Source}($x$)$ is defined (or the reverse, depending on your ordering convention).

For exercise 1, we took advantage of the fact we already knew $H$ was constructed from $G$ to ensure that our constructions worked. For exercise 2, we just have to check that we can prove they work just from the definition of algebraic groupoid.

I think you already have all of the necessary details -- I hope that this answer helps you organize them into answers to your questions and into understanding.


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