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I have some problems with characterizing equalizer: What are equalizer in a given poset P? Why is every monic an equalizer in the category of Sets? Is there a counterexample in another category?

Can someone help me? Thanks

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2 Answers 2

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Since in posets for every pair of objects $a$ and $b$ you have at most a morphism between them: this implies that for every pair $f,g \colon a \to b$ you have $f=g$ and so for every $h \colon x \to a$ clearly the equality $f\circ h = g \circ h$ holds. So every morphism to $a$ equalizes the pair $f$ and $g$ the equalizer should then be the a universal among the morphism which have $a$ as target, the identity $1_a$ do the trick, because for every morphism $h \colon x \to a$ clearly $x$ factors uniquely through $1_a$ as $x= 1_a \circ x$.

Equalizers are monic in every category. Here the proof. Let $f,g \colon a \to b$ be morphisms in a category $\mathbf C$ and an equalizer of $f$ and $g$, let's call it $e \colon c \to a$.

Suppose you have $x,y \in d \to c$ such that $e \circ x= e \circ y$ then clearly we have the equalities $$f \circ (e \circ x)= g \circ (e \circ x) = g \circ (e \circ y) = f \circ (e \circ y)$$ So the morphism $e \circ x/e \circ y$ equalize $f$ and $g$, but then this morphism must have a unique factorization through $e$ (which is an equalizer) and so $x=y$.

While is true that every equalizer is a monic the opposite doesn't always holds. For instance in posets every morphism is (trivially) monic but the only equalizers are the identities.

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  • $\begingroup$ i really mean monic implies equalizer (in sets) $\endgroup$
    – QEUO
    Commented Oct 14, 2013 at 11:54
  • $\begingroup$ @QEUO Ok, now I've made a change that should address the last part of the question :) $\endgroup$ Commented Oct 14, 2013 at 13:22
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Equalizers are defined for pairs of parallel morphisms. In a poset there is at most one morphism from here to yonder and thus no parallel morphisms, except for a morphism and itself. Any equalizer of the morphisms $f$ and $f$ is simply the identity morphism at the domain.

The other question relating to monics in $Set$ all being equalizers is answered by construction. If $f:A\to B$ is a monic function (i.e., it is injective) then consider the function $g:B\to B^*$, where $B^*$ is $B$ augmented with an extra element $*$, given by $g(b)=b$ if $b\in Im(f)$ and $f(b)=*$ otherwise. You can now show that $f$ is an equalizer of $g$ and the inclusion $B\to B^*$. This is in fact a special case of a much more general phenomenon, namely that in any topos monics and equalizers coincide (any equalizer is monic holds in any category).

It is certainly not the case that monos are equalizers in all categories. For one thing, equalizers do not need to exist, while a category may still have lots of monos. But even if all equalizers do exist, not all monos need be equalizers.

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    $\begingroup$ One may as well give an example where a monic is not an equalizer: in $\mathbf{Top}$, the identity function from a set with the discrete topology to the same set with a coarser topology is monic, but it is not an equalizer (since equalizers in $\mathbf{Top}$ carry subspace topologies, as discussed in a recent answer of mine). $\endgroup$
    – user43208
    Commented Oct 14, 2013 at 14:08

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