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I was studying the axioms of a category, and noted that one axiom says there is an element $1_X\in Hom(X,X)$ for any object $X$ which serves as the identity. Why is this axiom necessary? What happens if I drop this axiom?


Background: I can define the category of affine holomorphic symplectic varieties, by saying

  1. The objects are semisimple algebraic groups
  2. The morphisms $Hom(G,G')$ are affine holomorphic symplectic varieties with Hamiltonian $G\times G'$ action
  3. The composition of two morphisms $X\in Hom(G,G')$ and $Y\in Hom(G',G'')$ is given by the holomorphic symplectic quotient $X\times Y//G'$.

This becomes a nice symmetric monoidal category; the identity in $Hom(G,G)$ is $T^*G$.

Suppose I want to consider the category of hyperkähler manifolds instead. I can try the following

  1. The objects are semisimple compact groups
  2. The morphisms $Hom(G,G')$ are hyperkähler manifolds with Hamiltonian $G\times G'$ action
  3. The composition of two morphisms $X\in Hom(G,G')$ and $Y\in Hom(G',G'')$ is given by the hyperkähler quotient $X\times Y///G'$.

Now, the problem is that $T^*G_\mathbb{C}$ has a hyperkähler metric (constructed by Kronheimer) and almost acts like an identity, but not quite: given a hyperkähler manifold $X$ with $G$ action, $T^*G_\mathbb{C} \times X /// G$ is equivalent to $G$ as holomorphic symplectic varieties but not equivalent as hyperkähler manifolds.

What should I do?


For my purpose, I guess using the terminology semigroupoid would suffice (I just want to define the target "category" of a TQFT precisely.) But I'm curious what kind of hell will break loose if I drop this axiom, why the people who originally defined categories included this into the axiom, etc.

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It looks like you want to take the quotient by $G'$ in both cases. –  S. Carnahan Jun 15 '11 at 3:44
    
If you drop the axiom, you get a "category without identity" which is also called a semigroupoid. –  S. Carnahan Jun 15 '11 at 3:52
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Try ncatlab.org/nlab/show/semicategory instead. One would only have a semigroupoid if all arrows were invertible. And really, you can't do that, because you can't express invertibility, because you need identity arrows for that. As far as asking 'why identities', consider asking the same question for groups: why do groups have identity elements? –  David Roberts Jun 15 '11 at 4:42
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@David: wikipedia en.wikipedia.org/wiki/Semigroupoid says it doesn't have the invertibility axiom. Which is the standard definition of the terminology? –  Yuji Tachikawa Jun 15 '11 at 4:44
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Monoid= group without inverses. Semigroup= group without inverses and without identity. Category= many-object monoid or monoid-oid. So, a category without identity should be called "semigroup-oid". –  Qfwfq Jun 15 '11 at 13:11
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4 Answers

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Your structure can be described as a "category without identity", which has been given the names "semicategory" and "semigroupoid" presumably due to independent discoveries.

Some Googling suggests the term "semicategory" came first, in a 1972 TAMS paper by Mitchell. The name is motivated by applying an analogy connecting groups and semigroups to categories (as categories without identities or inverses), and it seems to be popular among people who study categories.

The term "semigroupoid" seems to have appeared first in Tilson, Categories as Algebra: an essential ingredient in the theory of monoids in Journal of Pure and Applied Algebra 48 (1987) 83-198. The name is motivated by applying an analogy connecting groups and groupoids to semigroups (as semigroups with multiple objects), and it seems to be popular among people who study semigroups.

I think the analogy is a bit weak on the semicategory side, since categories don't straightforwardly generalize groups. I'm not in charge, though.

John Baez points out in his TWF week 296 that there is a canonical way to make a category out of a semicategory, by formally adding, for each object, an identity element to the set of morphisms from that object to itself (and preserving all other morphism sets and composition laws). Any previously existing identities become idempotents. He notes that the categories formed this way are distinguished among all categories by the property that all invertible morphisms are identities. In particular, this process of formally adding identities is reversible in a canonical way, and no hell will break loose.

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Thanks Scott, TWF 296 was particularly useful. –  Yuji Tachikawa Jun 15 '11 at 15:01
    
I guess one can selectively add identities, so as to avoid those extra idempotents. –  Mariano Suárez-Alvarez Jun 15 '11 at 16:14
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I've never liked this "add an identity everywhere" approach. In the algebra case, this corresponds to the "one-point compactification", and a better approach is the "Stone-Cech compactification". For the latter, on the algebra side: if A is a possibly-non-unital algebra, you consider the algebra of endomorphisms of A-as-a-right-A-module. This is unital, and satisfies the right universal property. Something like this should be doable for categories as well. –  Theo Johnson-Freyd Jun 16 '11 at 4:55
    
@Theo: I must confess I do not understand your need to attach value judgments to definitions and constructions. On a more mathematical note, I can see how your "endomorphism monoid" method will work on each object, but I don't know how to make it into a functor from $SemiCat$ to $Cat$. –  S. Carnahan Jun 16 '11 at 5:38
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As Fernando points out, you can't talk about isomorphisms in a semicategory, which means that they won't be as much use as categories in describing universes of mathematical objects. But the category of semicategories has a surprisingly interesting relationship to that of categories. There is of course a forgetful functor $\mathrm{Cat} \to \mathrm{Semicat}$, and as Scott says it has a left adjoint that does what you expect. But it also has a right adjoint, which takes a semicategory S to the category of idempotents in S: the objects are idempotents $e \colon a \to a$ and a morphism $e \to e'$ is a morphism $f \colon a \to a'$ in S such that $fe = f = e'f$. So we get a monad on Cat whose unit is the canonical functor from a category to its idempotent-splitting completion, or Cauchy completion, or Karoubi envelope.

Böhm, Lack and Street use this framework here to talk about weak Hopf algebras. They show that 'weak monoids' fall naturally out of the formal theory of monads if instead of working directly in a bicategory you Cauchy-complete the hom-categories first.

Another application of semicategories and semifunctors is in computer science: Hayashi, Adjunction of semifunctors: categorical structures in nonextensional $\lambda$-calculus, TCS 41, shows how to describe $\lambda$-calculus without the $\eta$-law quite elegantly. I haven't worked it out, but it seems to me that this framework should also give a way of talking about 'weak limits' (the kind with not-necessarily-unique mediating morphisms) in terms of adjunctions.

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You mean a forgetful functor $\text{Cat} \to \text{Semicat}$. –  Qiaochu Yuan Jun 15 '11 at 16:27
    
Dammit, yes, thanks. Fixed. –  Finn Lawler Jun 15 '11 at 16:40
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+1 -- very interesting; I didn't know any of this! –  Todd Trimble Jun 15 '11 at 16:50
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One is often interested in isomorphisms in a category. After all, categories are seldom regarded as rigid objects, two categories are considered to be 'the same' if the are equivalent, and the notion of equivalence of categories relies on isomorphisms. Categories without identities are too rigid (at least from a categorical point of view).

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I suppose the extent to which hell breaks loose depends entirely on the purpose you're using categories for. Dropping identities presumably invalidates the Yoneda lemma, and therefore all the results in category theory that depend on it. But if you just want to use (monoidal) functors as bookkeeping devices without expecting "deeper" category theory to predict things for you, nothing much happens. To follow up on Scott's answer, there is for example a perfectly good theory about adjunctions when one ignores identities, and there is in fact a relation to formally adding identities. See Hayashi "Adjunction of semifunctors" in Theoretical Computer Science 41:95--104, 1985, and Hoofman and Moerdijk "A remark on the theory of semi-functors" in Mathematical Structures in Computer Science 5:1--8, 1995.

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Thank you for the reply. To talk about Yoneda's lemma, Hom needs to be a set, right? I'm not sure if the class of all hyperkahler manifolds is a set... I've heard that the class of all groups is not a set. ??? –  Yuji Tachikawa Jun 15 '11 at 19:12
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@Yuji Tachikawa As far as there is a bound on the cardinality of an atlas, then it is a set. For instance, the collection of compact hyperkahler manifolds is a set. The class of all groups is clearly not a set because each set has a group of bijections associated and these groups already form a class. –  Leo Alonso Jun 15 '11 at 21:20
    
@Leo Thanks, that was very clear. –  Yuji Tachikawa Jun 16 '11 at 0:41
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