Rel does not have all finite equalizers. I believe the smallest example is the parallel pair (1 1 0 0) and (0 0 1 1), seen as relations from the 4-element set to the 1-element set. Cheers, Jamie. 2009/8/20 Michael Barr <barr@math.mcgill.ca>:

On Thu, 20 Aug 2009, soloviev@irit.fr wrote:

...

To the list:

I am actually travelling (in Russia) and I need more or less  urgently a reference concerning pullbacks and pushouts in the category of sets and relations - do they always exist etc - I would not adress it to the list if it would be not urgent and I would not have some difficulty with search from here -

Best to all -

Sergei Soloviev

You should check this, but it seems right.  Rel is self dual and each object is too.  So limits are colimits (of the dual diagram).  Rel has arbitrary sums and products--they are disjoint unions.  The empty set is initial and terminal.  So to have pullbacks you need equalizers.  So let A and B be sets and R,S \inc A x B.  Let A_0 be the subset of A consisting of all a such that (a,b) \in R iff (a,b) \in S.  Then it seems to me that the inclusion function of A_0 into A is the equalizer of R and S.

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Rel doesn't have equalizers (pullbacks, pushouts, or co-equalizers). It's rather surprising, and apparently not very well known. Counter-example. Let A = { a, b }, write id for the identity A -> A, and define z : A -> A by adding an edge to id: a --- a / / / b --- b Then id and z do not have an equalizer in Rel. To see why, first note that we can identify any equalizer with a set Q of non-empty subsets of A, on each of which z and id have the same direct image [Footnote 1]. The only such candidate subsets are { a } { a, b } This leaves us with just 2x2 = 4 possibilities for our equalizer, all of which fail [Footnote 2]. Interestingly, the related self-dual category Coh of coherence spaces *does* have equalizers (generalizing co-equalizers/pushouts of partial functions). A few years back I conjectured that Coh should have pullbacks/equalizers, based on thinking about path-based composition in categories of linkings (Kelly-Mac Lane graphs, proof nets, etc). Robin Houston worked out the details of the Coh definition. See this draft for the use of pullbacks for linking composition: http://arxiv.org/PS_cache/arxiv/pdf/0805/0805.1441v1.pdf Dominic Footnote 1. Let e : Q --> A be an equalizer. Were the direct image e[q] of q empty in A, then uniqueness in the universal property would fail. Were q,r in Q distinct with the same direct image e[q]=e[r] \subseteq A, then again uniqueness would fail. Thus we can identify an equalizer with a set of distinct non-empty subsets of A. For any such subset q, we must have the direct images z[q] and id[q] equal subsets of A, otherwise we would not have id e = z e . Footnote 2. Let e : Q --> A be our supposed equalizer. Case 1: Q empty. Consider the test function f : X --> A where X = {x} and f is a single edge x---a . Then for no f' : X --> Q can we have f = e f' as required (since the e is empty, while f is not). Case 2: Q = { a,b }. Consider the same test function as in case 1. For no f' : X --> Q can we have f = e f' (since f hits a but not b, while e f' hits neither (if f' is empty) or both (if f' has an edge)). Case 3: Q = { a }. Consider the test function g : X --> A where X = {x} and g has two edges x---a and x---b . For no g' : X --> Q can we have g = e g' (since g hits both a and b, while e hits only a). Case 4: Q contains { a,b } and { a }. Consider the same test function as in case 3. Since g hits both a and b, we must have an edge in g' from x to { a,b }. Then an edge in g' from x to { a } is optional, while still having g = e g', so uniqueness fails in the universal property. On Thu, 20 Aug 2009, Michael Barr wrote:

On Thu, 20 Aug 2009, soloviev@irit.fr wrote:

...

To the list:

I am actually travelling (in Russia) and I need more or less urgently a reference concerning pullbacks and pushouts in the category of sets and relations - do they always exist etc - I would not adress it to the list if it would be not urgent and I would not have some difficulty with search from here -

Best to all -

Sergei Soloviev

You should check this, but it seems right. Rel is self dual and each object is too. So limits are colimits (of the dual diagram). Rel has arbitrary sums and products--they are disjoint unions. The empty set is initial and terminal. So to have pullbacks you need equalizers. So let A and B be sets and R,S \inc A x B. Let A_0 be the subset of A consisting of all a such that (a,b) \in R iff (a,b) \in S. Then it seems to me that the inclusion function of A_0 into A is the equalizer of R and S.

[For admin and other information see: http://www.mta.ca/~cat-dist/ ]

Dear all,

let A and B be sets and R,S \inc A x B. Let A_0 be the subset of A consisting of all a such that (a,b) \in R iff (a,b) \in S. Then it seems to me that the inclusion function of A_0 into A is the equalizer of R and S.

I already answered Sergei privately, but this should be corrected publicly: Rel does not have equalisers. In fact, it has almost no equalisers. For a counterexample, consider the sets A = {0,1} and B = {0}, and the parallel relations R = A x B and S = {(0,0)}. Their equaliser, if it existed, must be contained in T = {(0,0)}. But T' = {0} x A also satisfies RT'=ST', but does not factor through any subrelation of T. Best wishes, Chris Heunen [For admin and other information see: http://www.mta.ca/~cat-dist/ ]