# Projection map being a closed map

Let $\pi: X \times Y \to X$ be the projection map where $Y$ is compact. Prove that $\pi$ is a closed map.

• First I would like to see a proof of this claim.

• I want to know that here why compactness is necessary or do we have any other weaker condition other than compactness for the same result to hold.

Suppose $$Z⊂X×YZ \subset X \times Y$$ is closed, and suppose $$x0∈X∖π[Z]x_0 \in X \setminus \pi[Z]$$.
For any $$y∈Y,(x0,y)∉Zy \in Y, (x_0, y) \notin Z$$, and as $$ZZ$$ is closed we find a basic open subset $$U(y)×V(y)U(y) \times V(y)$$ of $$X×YX \times Y$$ that contains $$(x0,y)(x_0, y)$$ and misses $$ZZ$$.
The $$V(y)V(y)$$ cover $$YY$$, so finitely many of them cover $$YY$$ by compactness, say $$V(y1),…,V(yn)V(y_1),\ldots,V(y_n)$$ do. Now define $$U=∩ni=1U(yi)U = \cap_{i=1}^{n} U(y_i)$$, and note that $$UU$$ is an open neighbourhood of $$x0x_0$$ that misses $$π[Z]\pi[Z]$$: suppose that there is some $$(x,y)∈Z(x,y)\in Z$$ with $$π(x,y)=x∈U\pi(x,y) = x \in U$$. Then $$y∈V(yi)y \in V(y_i)$$ for some $$ii$$, and as $$x∈U⊆U(yi)x \in U \subseteq U(y_i)$$ (as $$UU$$ is the intersection of all $$U(yi)U(y_i)$$) we get that $$(x,y)∈(U(yi)×V(yi))∩Z(x,y) \in (U(y_i) \times V(y_i)) \cap Z$$ which contradicts how these sets were chosen to be disjoint from $$ZZ$$. So $$U∩π[Z]=∅U \cap \pi[Z]=\emptyset$$ and $$π[Z]\pi[Z]$$ is closed.
To see that the closed projection property implies compactness: suppose $$XX$$ has the closed projection property along $$XX$$, and let $$F\cal{F}$$ be a filter on $$XX$$. Define a space $$YY$$ that is as a set $$X∪{∗},∗∉XX \cup \{\ast\}, \ast \notin X$$, where $$XX$$ has the discrete topology and a neighbourhood of $$∗\ast$$ is of the form $$A∪{∗}A \cup \{\ast\}$$ with $$A∈FA \in \cal{F}$$. Then $$D={(x,x):x∈X}D = \{(x,x): x \in X\}$$ is a subset $$X×YX \times Y$$ and closedness of the projection $$p:X×Y→Yp: X \times Y \rightarrow Y$$ implies that some point $$(x,∗)(x,\ast)$$ is in its closure: $$DD$$ cannot be closed in $$X×YX \times Y$$, because $$p[D]=Xp[D] = X$$ is not closed in $$YY$$, as $$∗∈¯p[D]=p[¯D]\ast \in \overline{p[D]} = p[\overline{D}]$$, by closedness (and continuity) of $$pp$$. This $$xx$$ is an adherence point of the filter, because if $$A∈FA \in \mathcal{F}$$ and $$x∈Ox \in O$$ where $$OO$$ is open in $$XX$$, then $$O×(A∪{∗})O \times (A \cup \{\ast\})$$ is basic open in $$X×YX \times Y$$ and so intersects $$DD$$ in some $$(p,p)(p,p)$$, $$p∈Xp \in X$$. This $$p∈O∩A≠∅p \in O \cap A \neq \emptyset$$, showing that $$xx$$ is an adherence point of $$F\mathcal{F}$$.