The pushout in the category of topological spaces is given by the gluing of spaces along continuous maps. Does there exist a similar "easy" topological description of the pullback? Does it even always exist in this category?


1 Answer 1


Yes, exists, and is a subspace of $A\times B$, in case you want to pull back $f:A\to C$ and $g:B\to C$. Namely, a kind of 'equalizer': $$A{}_f\times_g B:= \{ (a,b)\mid f(a)=g(b)\} $$ I think you can draw it somehow..

Somehow.. generally, pullback 'asks' for the solution (when $f=g$?), and pushout 'realizes' or 'forces' (topologically: 'glues') the solution, saying: "let $f(a)=g(a)$ in the new space for all $a$".

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    $\begingroup$ It is precisely an equalizer. In any category with finite products, pullbacks can be computed by computing a corresponding equalizer. Note also that the forgetful functor $\text{Top} \to \text{Set}$ has both a left and a right adjoint, hence preserves both limits and colimits, so in particular on underlying sets the pullback in $\text{Top}$ is just the pullback in $\text{Set}$, and it's straightforward to figure out what the required topology is. $\endgroup$ Oct 17, 2012 at 17:57

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