We discuss elementary properties of the pushout diagram in $\mathsf{Top}$.
Notice that a subset $A\subset Y\underset{X}\sqcup Z$ is open or closed iff $\widetilde{f}^{-1}(A)$ and $\widetilde{g}^{-1}(A)$ are.
We also observe that
\[\widetilde{g}^{-1}(\operatorname{im}(\widetilde{f}))=\operatorname{im}(f).\]Proposition 1. If $\operatorname{im}(f)$ is closed (resp. open) in $Y$, then
$\widetilde{g}$ restricted to $Y\setminus\operatorname{im}(f)$ is an open (resp. closed) embedding whose image is $Y\underset{X}\sqcup Z\setminus\operatorname{im}(\widetilde{f})$.
Therefore, $\operatorname{im}(\widetilde{f})$ is also closed (resp. open).
Proof. Almost trivial using the observations above.
Proposition 2. If $f$ is surjective (or injective), then so is $\widetilde f$.
Proof. This is true in $\mathsf{Set}$.
Proposition 3. If $f$ is a quotient map, then so is $\widetilde f$.
Proof. This follows easily from the universal properties. More specifically, we are given a function $\varphi:Y\underset{X}\sqcup Z\to P$ such that $\varphi\widetilde f$ is continuous. Since $\widetilde f$ is surjective, $\varphi$ is the unique map making this triangle (see figure below) commute. Therefore $\varphi$ is the same as the map induced by $f$ and the pushout, hence is continuous.
Proposition 4. If $f$ is an embedding, then so is $\widetilde f$.
Proof. We need that the induced bijection $\widetilde f:Z\to\operatorname{im}\widetilde f$ is an open map. Fix an open subset $U\subset Z$.
Since $f$ is an embedding, there is some open $V\subset Y$ such that $f^{-1}(V)=g^{-1}(U)$. We thus have the commutative diagram (see figure below) which induces a map $1_W:Y\underset{X}\sqcup Z\to\Sigma$, where $\Sigma$ is the Sierpiński space. In other words, $W$ is open and satisfies $\widetilde f^{-1}(W)=U$.
Corollary 5. If $f$ is a closed (or open) embedding, then so is $\tilde f$.
When $f$ is a closed embedding, the pushout is often called the adjunction space, denoted as $Z\cup_fY$.
This notion is very useful and encompasses many common constructions, for example,
Proposition 6. If $f$ is a closed embedding, and $X$, $Y$, $Z$ are T1 (or T4), then $Y\underset{X}\sqcup Z$ is T1 (or T41) as well.
Proof. Suppose that $Y$, $Z$ are T1. In view of Proposition 1, any points in $Y\underset{X}\sqcup Z\setminus\operatorname{im}(\widetilde f)$ is closed. On the other hand, $\widetilde{g}^{-1}(\widetilde{f}(z))=f(g^{-1}(z))$ is closed, and $\widetilde{f}^{-1}(\widetilde{f}(z))=z$ by injectivity. This proves the first statement.
Building on this result, we apply (the converse of) Tietze extension theorem for the second part. Fix a map $\varphi_0:V\to\mathbb{R}$, where $V\subset Y\underset{X}\sqcup Z$ is a closed subset. We look for an extension of this, using the universal property, of course.
Let $\varphi_1:Z\to\mathbb{R}$ be an extension of $\varphi_0\widetilde{f}:\widetilde{f}^{-1}(V)\to\mathbb{R}$. Viewing $Z$ as a subset of $Y\underset{X}\sqcup Z$, we can glue $\varphi_0$ and $\varphi_1$ together to obtain $\varphi:V\cup Z\to\mathbb{R}$.
We can then pull this back along $\widetilde{g}$ and extend it to $\psi:Y\to\mathbb{R}$. It’s now clear that $\varphi_1g=\psi f$, which concludes the proof. Indeed,
\[\psi f=\varphi\widetilde{g}f=\varphi\widetilde{f}g=\varphi_1g.\]
Unfortunately, the above does not hold for T2 spaces. Take $Z=(2,\text{discrete})$ and let $X=V_1\cup V_2$ be the union of two disjoint closed subsets of $Y$. Hausdorff property of the adjunction space (which is just a quotient of $Y$ which glues $V_1$ and $V_2$ into two points) implies that $V_1$ and $V_2$ can be separated by open neighborhoods. However, there are Hausdorff spaces that are not even regular, e.g. this space.
Proposition 7. If $f$ is a cofibration, then so is $\widetilde{f}$.
Proof. Easy to see from the HEP diagram. Omitted.
Post date: 2025/02/23
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T4 means T4 and T1. ↩