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Suppose I have a category C. And there are two objects X and Y with no morphisms between them. I've checked up "quotient category" on wikipedia, but there I can only make isomorphic objects with morphisms between them.

Is there a categorial notion available that I can use in this case?

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  • $\begingroup$ It seems to be a bit weird. Then you would need a definition of what a composition of a morphism $C\rightarrow X$ and a morphism $Y\rightarrow C'$ is. Seems to be a bit strange to me. $\endgroup$
    – HenrikRüping
    Commented Aug 12, 2010 at 10:58
  • $\begingroup$ There is a way to consider a category to be defined only by morphisms - with the identity morphisms taking on the roles of the objects. Does this help making the quotient approach work for you? $\endgroup$
    – Mikael Vejdemo-Johansson
    Commented Aug 12, 2010 at 11:15
  • $\begingroup$ Not sure I'm correctly understanding your question: do you want to describe a "quotient-like" category D together with a functor $F:C\to D$ such that $F(X)\cong F(Y)$ and which is universal with respect to this property? $\endgroup$
    – domenico fiorenza
    Commented Aug 12, 2010 at 12:43
  • $\begingroup$ I am afraid that I do not understand the question. What do you want? It would be easier if you wanted no chain of morphisms between them, then the idea of component would do the trick. Do you want a test to show that there are no morphisms between two specific objects? $\endgroup$
    – Tim Porter
    Commented Aug 12, 2010 at 12:44
  • $\begingroup$ If, as domenico suggests, you are looking for the universal functor C -> D such that the images of X and Y are isomorphic, you should be able to describe it via generators and relations if you admit large categories (i.e. categories whose hom's can be classes instead of sets). If you don't like this, the large amount of literature on localisation bears witness to the difficulty of the general problem. Do you have a particular category or class of categories in mind? $\endgroup$
    – anon
    Commented Aug 12, 2010 at 12:49

2 Answers 2

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Short answer: yes. There is a special class of 2-categorical limits called iso-inserters that does the trick. The paper to check is M. Kelly's "Elementary observations on 2-categorical limits", Bull. Austr. Math. Soc. 39 (1989), 301-317.

Instead of explaining what these are let me go about explaining how you would make two objects $a$ and $b$ isomorphic. First pass to the underlying graph of the category and insert two edges $a\to b$ and $b\to a$ then take the free category of this new graph. You have a graph morphism from the original category into this new graph. Quotient the category to force the graph morphism to be a functor. Now quotient the category again to make the edges you inserted to be mutually inverse. There is only a small snag to this construction: the category you end up may not be locally small, because inserting isomorphisms may create a proper class of new morphisms. This is where things like "calculus of fractions" come in.

Hope it helps, regards, G. Rodrigues

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  • $\begingroup$ More generally, categories can be seen as "just another (essentially) algebraic theory": just as you can always freely adjoin elements to a group, or a ring, or "adjoin equations" (i.e. by quotienting), similarly you can always adjoin an object, a morphism, or an equation (or a bunch of any of these) to a category. As G.R. points out, this has slight size issues: as with groups etc., the size of the whole structure won't blow up much (at most countably much), so a small cat will remain small, and so on; but local smallness may not be preserved. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Aug 12, 2010 at 15:32
  • $\begingroup$ (cont'd) but as anon's comment on the OP points out, the resulting categories are often very difficult to get a handle on (essentially because many new composites can arise, even of arrows that already existed). $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Aug 12, 2010 at 15:35
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Domenicos comment lead to the following idea (I am posting this as an answer, as it is too long for a comment):

Let $CAT$ denote the category of small categories and $CAT'$ denote the category, whose objects are small categories except for the fact that the composition needn't be defined on the whole of $Mor(A,B)\times Mor(B,C)$ (but just on a subset of it). Associativity and so on should hold, whenever it is defined.

Then there is a obvious inclusion Functor $CAT\rightarrow CAT'$. One should check, whether it has a left adjoint $L:CAT'\rightarrow CAT$. Then one can make out of the data above a object in CAT' by adding an additional isomorphism from $X$ to $Y$ and one doesn't have to worry aboutthe compositions of that iso with the morphisms in CAT. Using $L$ one could make a honest category out of this.

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  • $\begingroup$ By left inverse, surely you mean left adjoint. $\endgroup$
    – Saul Glasman
    Commented Aug 12, 2010 at 13:12
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    $\begingroup$ But unfortunately, I think I can prove that such an adjoint cannot exist. A functor admitting a left adjoint must preserve limits; I claim that the inclusion $CAT \to CAT'$ does not commute with the product of $C_2$, the group with two elements, with itself. In $CAT'$, this product is a one-object category with three morphisms: the identity and two non-composable automorphisms of order $2$. But in $CAT$, I believe, this product is the group $C_2 \times C_2$, which has four morphisms. $\endgroup$
    – Saul Glasman
    Commented Aug 12, 2010 at 13:30
  • $\begingroup$ agreed. I will delete my answer soon (maybe tomorrow). $\endgroup$
    – HenrikRüping
    Commented Aug 12, 2010 at 14:03
  • $\begingroup$ Actually, now I think I was wrong. $C_2 \times C_2$ doesn't actually factor through the object I claimed was the product in $CAT'$ - so maybe the products do agree. $\endgroup$
    – Saul Glasman
    Commented Aug 12, 2010 at 14:43

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