On equalizers in Top 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?
 A: 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. 
A: 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.
