# Show that the Quotient map is closed.

Let $X$ and $Y$ be topological spaces. Let $X \cup Y$ be the space with disjoint union topology.Suppose $A \subset X, A$ closed, $f : A \to Y$ be a continuous closed map.Consider the quotient map $\pi: X \cup Y \to X \cup _{f} Y$.Then $\pi$ is closed map.

I am trying to prove using the definitions directly but its getting bit messy.Is there any 'smart' way to prove $\pi$ is closed?

• If $F \subset X \cup Y$ is closed, a) what does that mean, and b) what is $\pi^{-1}(\pi(F))$? Commented Sep 5, 2016 at 15:32

In the following diagram let the map $A \hookrightarrow X$ and the maps to $X \cup Y$ be the inclusions. The maps $\bar f$ and $j: Y \hookrightarrow X \cup_f Y$ send a point to its equivalence class, so the outer rectangle and the triangles commute. Note that closedness of $\pi$ is equivalent to closedness of both $\bar f$ and $j$, so showing that each of these two maps is closed is maybe less messy.

Consider closed sets $B \subseteq X$ and $C\subseteq Y$. You need to show that $\pi^{-1}(\bar f(B))$ and $\pi^{-1}(j(C))$ are closed in $X \cup Y$. For either set, there exists a simple formula from which you should see, using closedness of $A$ and of $f$ whenever needed, that the set is closed. Can you find them? Let me know if you struggle and I'll provide some hints.

• Sorry..I really don't see why the claimed sets are closed probably because my knowledge to ad junction spaces is very limited.Could you please explain more? Commented Sep 5, 2016 at 17:43
• You have $\pi^{-1}(\bar f(B)) = B \cup f(B\cap A) \cup f^{-1}(f(B\cap A))$. Do you see why this set is closed? For $\pi^{-1}(j(C))$ there is a similar formula @VictorBarg Commented Sep 5, 2016 at 18:04
• Using the continuity of $f$ and Closedness of $f$ its clear to me that the set is closed.But i dont actually see how do you get that formula ?Sorry for troubling you. Commented Sep 5, 2016 at 18:39
• @RafaelHolanda: Using the package tikz in Latex. Here is the code pastebin.com/y2HWf50r Commented Sep 6, 2016 at 11:23
• You are welcome. And don't be ashamed of your questions, it is normal to have quite some problems when you first learn a topic. Don't forget to also figure out the formula for $\pi^{-1}(j(C))$, which is actually simpler than the formula for $\pi^{-1}(\bar f(B))$. @VictorBarg Commented Sep 6, 2016 at 11:54

I would just follow the definitions. Suppose that $F\subseteq X\cup Y$ is closed. Then $F$ is the disjoint union of the sets $F_0=(F\cap X)\setminus A$, $F_1=F\cap A$, $F_2=(F\cap Y)\setminus f[A]$, and $F_3=F\cap f[A]$.

• Show that $\pi^{-1}[\{\pi(z)\}]=\{z\}$ if $z\in F_0\cup F_2$.
• Show that $\pi^{-1}[\{\pi(z)\}]=f^{-1}[\{f(z)\}]\cup\{f(z)\}$ if $z\in F_1$.
• Show that $\pi^{-1}[\{\pi(z)\}]=f^{-1}[\{z\}]\cup\{z\}$ if $z\in F_3$.
• Conclude that $\pi^{-1}\big[\pi[F]\big]=F\cup f^{-1}\big[f[F\cap X]\big]\cup f^{-1}[F\cap Y]$, which is closed (why?).
• Thank you..Could you please explain why $F$ is union of $F_i$'s ? I know that a set $F$ in $X \cup Y$ is closed iff $F \cap X$ and $F \cap Y$ is closed in $X$ and $Y$ respectively. Commented Sep 6, 2016 at 4:16
• @Victor: Every point of $X\cup Y$ belongs to exactly one of the four sets $A$, $X\setminus A$, $f[A]$, and $Y\setminus f[A]$; I’ve simply applied that to $F$ by intersecting each of those four sets with $F$. Commented Sep 6, 2016 at 4:26
• Yeah.Just now i also realized that.Btw,it it true that closed set in $X \cup Y$ is the disjoint union of $B_1$ and $B_2$ where $B_1$ and $B_2$ are closed in $X$ and $Y$ respectively ? Commented Sep 6, 2016 at 4:28
• @Victor: Yes, it is. Suppose that $F$ is closed in $X\cup Y$. Each of the sets $X$ and $Y$ is closed in $X\cup Y$, so $F\cap X$ and $F\cap Y$ are closed in $X\cup Y$, and clearly $F$ is the disjoint union of these two sets. Commented Sep 6, 2016 at 4:29
• @Victor: You’re welcome; glad to help! Commented Sep 6, 2016 at 16:38