# Pushouts preserve homeomorphisms

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.

• Are you able to show that $B$ satisfies the universal property of a pushout? Oct 18, 2023 at 8:07
• 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 Oct 18, 2023 at 8:28
• 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 Oct 18, 2023 at 9:05
• What I mean is, that your diagram for $D = B$, $g = \mathrm{id_B}$ and $k = h\circ f^{-1}$ is a pushout diagram. Oct 18, 2023 at 10:10
• math.stackexchange.com/questions/4160501/… Oct 18, 2023 at 14:21

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 $$C$$.
• 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 Oct 18, 2023 at 20:41
• @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 $f,k,g,h,A,B,C,D$. 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.