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Wikipedia says "The equalizer of a pair of morphisms is given by placing the subspace topology on the set-theoretic equalizer." for the category $\mathbf{Top}$.

What is the simplest way to prove this? It seems to be an instance of a more general (not only about $\mathbf{Top}$) theorem. Isn't it?

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If you have a category whose objects are sets and whose morphisms are functions of sets, the equalizer will just be the set-theoretic equalizer, if it exists. – Joe Johnson 126 Oct 14 '13 at 12:43
@JoeJohnson126: But we have topospaces as objects, not sets. How can your comment help in this case? Maybe there is some "embedding"? – porton Oct 14 '13 at 12:46
@Joe: You assume the forgetful functor preserves limits. – Hurkyl Oct 14 '13 at 12:46
@porton By "objects are sets", I mean that the objects are sets with, perhaps, extra structure, like a topology. Also, Hurkyl is right, forgetful functors must preserve limits as well. – Joe Johnson 126 Oct 14 '13 at 12:57
@JoeJohnson126 A counterexample to your claim is the category of coalgebras over some field. Here the underlying-set functor does not preserve equalizers (which do in fact exist). Interestingly, this is a case where the underlying-set functor has a right adjoint, not a left adjoint. – user43208 Oct 14 '13 at 13:08
up vote 4 down vote accepted

Since the underlying-set functor $\hom(1, -): \mathbf{Top} \to \mathbf{Set}$ preserves all limits (being a representable functor), it preserves equalizers in particular. So we know that the underlying set of the equalizer must be the equalizer as computed in $\mathbf{Set}$.

The only question then is what is the correct topology on the equalizer $i: E \to A$ (of a pair of arrows $f, g$ from $A$ to $B$ say). We know that $i$ must be continuous, and this means that $i^{-1}(U)$ must be open for every open $U \subseteq A$; that is, thinking of $i$ as an inclusion, we must have $U \cap E$ open in $E$. So at least the correct topology must contain the subspace topology. On the other hand, if we consider the inclusion map $j: E_{sub} \to A$ where $E_{sub}$ is the underlying set equipped with the subspace topology, then surely $f j = g j$, so this would have to factor through the correct topology, meaning the correct topology must be contained in the subspace topology. So it must be the subspace topology.

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Sorry for a stupid question: why is it representable and equal to hom(1,−)? – porton Oct 14 '13 at 12:57
The set of continuous maps $1 \to X$ from a one-point space $1$ is the same as the set of points of $X$, and the underlying function of a continuous map $f: X \to Y$ is determined by what it does to points $1 \to X$. So the underlying-set functor can be identified with $\hom(1, -)$, which is representable by definition. – user43208 Oct 14 '13 at 13:02

The forgetful functor is both a left and a right adjoint (the other side of the adjunction is the indiscrete and discrete topology respectively).

Because it is a right adjoint, it preserves all limits. In particular, it preserves equalizers.

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But this does not tell us which topology on the set-theoretic equalizer is the right one, does it? – Rasmus Oct 14 '13 at 12:47
How to prove (desirably for many concrete categories not only $\mathbf{Top}$) that it has adjoints? – porton Oct 14 '13 at 12:48
Have a look at, especially page 362. – drhab Oct 14 '13 at 13:09
You can find this and other marvels in the "Joy of Cats" book, as I suggested in a previous question. Follow @drhab link. – magma Oct 14 '13 at 13:37
IIRC means: "If I recall correctly". If I recall correctly. Right? :-) – magma Oct 14 '13 at 13:40

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