The pushout of an open/closed injective map is open/closed In this question, it is proved that the pushout of an open embedding is still open. I have two question about this:


*

*The given answer is very set-theoretic in spirit – is there a more category-theoretical approach to proving this, rather using universal properties instead of the set realization of pushouts?

*Does this statement hold true for closed maps as well?
 A: Ok, after thinking about it for a bit, I found an answer myself:
Say there’s a pushout square in $\mathrm{Set}$ or $\mathrm{Top}$:
\begin{align*}
 \require{AMScd}
 \begin{CD}
  T @>{f}>> A\\
  @V{g}VV @VV{α}V \\
  B @>>{β}> X
 \end{CD}
\end{align*}


*

*In $\mathrm{Set}$: If $f$ is injective, then so is $β$:
For any set $C ⊂ B$, by the injectivity of $f$ the set $D = f(g^{-1}(C))$ is disjoint from $f(g^{-1}(B\setminus C))$ and so the indicator maps $1_C \colon B → \{0,1\}$ and $1_D\colon A → \{0,1\}$ satisfy
$$1_C ∘ g = 1_{g^{-1}(C)} = 1_{f^{-1}(D)} = 1_D ∘ f,$$
because $f^{-1}(D) = f^{-1}(f(g^{-1}(C)) = g^{-1}(C)$ by the injectivity of $f$.
Therefore, by the universal property of pushouts there’s a $χ \colon X → \{0,1\}$, such that $1_C = χ ∘ β$. Thinking about this for a second, this implies that $β(C)$ must be disjoint from $β(B\setminus C)$. As $C$ has been arbitrarty, this implies $β$ must be injective.

*In $\mathrm{Set}$: $E = α^{-1}(β(B)) ⊂ f(T)$:
Use the universal property of pushouts on $1_{f(T)}\colon A → \{0,1\}$ and $1_B \colon B → \{0,1\}$ which obviously agree on $T$ via $f$ and $g$ respectively. Then there’s a $χ\colon X → \{0,1\}$ such that $χα = 1_{f(T)}$ and $χβ = 1_B$. Then
$$1_{f(T)}(E) = χα(E) = χα(α^{-1}(β(B))) ⊂ χβ(B) = 1_B(B) = 1,$$ and so $E ⊂ f(T)$.

*In $\mathrm{Top}$: If $f$ is injective and open/closed, then so is $β$:
By (1.), $β$ is injective (as pushouts in $\mathrm{Top}$ are automatically pushouts in $\mathrm{Set}$ as well). Let $C ⊂ B$ be open/closed. Then


*

*$β^{-1}(β(C)) = C$, by $β$ being injective, hence is open/closed.

*Let $D = α^{-1}(β(C)))$. Then $f^{-1}(D) = g^{-1}(C)$ is open/closed in $T$ and so is $f(f^{-1}(D))$ because $f$ is open/closed. But $f(f^{-1}(D)) = D$, because of (2.).



So yes, the statement generalizes to any property which $f$ preserves under images and preimages and which is characterized at $X$ through the preimages of $α$ and $β$. The reasoning here is a bit more categorical in flavour, but I guess one can use only subobject-classifier-parlance to prove the results as well.
