Consider the following diagram in $\mathbf{Top}$: \begin{array} AA & \stackrel{h}{\longrightarrow} & B \\ \downarrow{f} & & \downarrow{g} \\ C & \stackrel{k}{\longrightarrow} & D \end{array} and suppose that it is a pushout. How can I show that $f$ being an homeomorphism implies that $g$ must be also?

I think that this boils down to pushouts preserving isomorphisms, but I couldn't find a proof for it.

  • $\begingroup$ Are you able to show that $B$ satisfies the universal property of a pushout? $\endgroup$
    – Claudius
    Oct 18, 2023 at 8:07
  • $\begingroup$ Can you elaborate what it means for $B$ to satisfy the universal property? I know that this diagram satisfies the universal property since it was given to be a pushout. @Claudius $\endgroup$ Oct 18, 2023 at 8:28
  • $\begingroup$ Do you mean that this diagram is a pushout? $$ \begin{array} AA & \stackrel{h}{\longrightarrow} & B \\ \uparrow{f^{-1}} & & \uparrow{h \circ f^{-1}} \\ C & \stackrel{\operatorname{id}}{\longrightarrow} & C \end{array} $$ @ronno $\endgroup$ Oct 18, 2023 at 9:05
  • $\begingroup$ What I mean is, that your diagram for $D = B$, $g = \mathrm{id_B}$ and $k = h\circ f^{-1}$ is a pushout diagram. $\endgroup$
    – Claudius
    Oct 18, 2023 at 10:10
  • 1
    $\begingroup$ math.stackexchange.com/questions/4160501/… $\endgroup$
    – Julián
    Oct 18, 2023 at 14:21

2 Answers 2


To clean up the comments:

The maps $hf^{-1}:C\to B,1_B:B\to B$ induce (why?) a map $\lambda:D\to B$ which in particular satisfies $\lambda g=1_B$ (existence). To show $g\lambda=1_D$, we use the uniqueness part of the universal property of a pushout square. Simply check that $g\lambda$ and $1_D$ have the same components when ‘restricted’ to $B$ and $C$.

This works in any category. Exercise: do the same and show a pullback of an isomorphism is again an isomorphism. Prove this directly and prove this as a corollary of the result for pushouts.

  • $\begingroup$ So $\lambda$ is induced by the fact that, then $B$ is a pushout of the diagram $A \xleftarrow{f^{-1}} C \xrightarrow{hf^{-1}} B$? @FShrike $\endgroup$ Oct 18, 2023 at 20:41
  • $\begingroup$ @NathanielJohonson In my setup $\lambda$ has domain $D$, not $B$, so a pushout with tip $B$ would not be relevant. $\lambda$ is induced from the pushout square the question gives you, the one with $f,k,g,h,A,B,C,D$. $\endgroup$
    – FShrike
    Oct 19, 2023 at 8:14

A little bit more abstractly, this is because pushout is a left adjoint. For convenience, consider the dual statement, i.e. that pullback is a right adjoint. More precisely, for any $f:A\to B$, taking the pullback along $f$ is a functor from the slice category $C/B$ to the slice $C/A$. This functor has always as left adjoint the composition with $f$. From this, since right adjoint preserve limits, taking the pullback along $f$ preserves terminal objects in the slice, but in a slice category an object is terminal iff it is an iso in the base category, thus the pullback of an iso is iso. Your statement then follows by doing this proof in C^op.


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