I've read that a continuous $f:X\rightarrow S$ is proper (inverse images of compacts are compacts) iff for all other continuous maps $g:Y\rightarrow S$ the projection $X\times _SY\rightarrow Y$ in the pullback square below (in $\mathsf{Top}$) is a closed map. $$\require{AMScd} \begin{CD} X\times _S Y @>>> Y\\ @VVV @VV{g}V\\ X @>>{f}> S \end{CD}$$

  1. How does one prove this?
  2. Does it fail if we drop the assmptions $f,g$ are continuous? I.e we let the pullback be in $\mathsf{Set}$, with the pullback still topologized as a subspace of the product.
  • $\begingroup$ This is in Top? What's your definition of "proper" (inverse image of compact is compact, or something else)? What's the definition of the subscripted product? $\endgroup$ – Henno Brandsma Jan 23 '16 at 13:56
  • $\begingroup$ What's your definition of proper map if not this? $\endgroup$ – Zhen Lin Jan 23 '16 at 13:57
  • $\begingroup$ @HennoBrandsma, Zhen, I editted the question to clarify $\endgroup$ – user153312 Jan 23 '16 at 14:01
  • 3
    $\begingroup$ (1) The proof is not obvious even in the special case where $S$ is a point. It involves some carefully chosen topological spaces. See tag 005M in [Stacks]. $\endgroup$ – Zhen Lin Jan 23 '16 at 17:15
  • $\begingroup$ Maps with this property are called universally closed. It turns out that universally closed maps are precisely the perfect maps (closed maps with compact fibers, sometimes called proper maps), as proven in the page linked by Zhen Lin. $\endgroup$ – Stefan Hamcke Jan 29 '16 at 3:03

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