Let π:X×Y→X be the projection map where Y is compact. Prove that π 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.

**Answer**

Suppose Z⊂X×Y is closed, and suppose x0∈X∖π[Z].

For any y∈Y,(x0,y)∉Z, and as Z is closed we find a basic open subset U(y)×V(y) of X×Y that contains (x0,y) and misses Z.

The V(y) cover Y, so finitely many of them cover Y by compactness, say V(y1),…,V(yn) do. Now define U=∩ni=1U(yi), and note that U is an open neighbourhood of x0 that misses π[Z]: suppose that there is some (x,y)∈Z with π(x,y)=x∈U. Then y∈V(yi) for some i, and as x∈U⊆U(yi) (as U is the intersection of all U(yi)) we get that (x,y)∈(U(yi)×V(yi))∩Z which contradicts how these sets were chosen to be disjoint from Z. So U∩π[Z]=∅ and π[Z] is closed.

To see that the closed projection property implies compactness: suppose X has the closed projection property along X, and let F be a filter on X. Define a space Y that is as a set X∪{∗},∗∉X, where X has the discrete topology and a neighbourhood of ∗ is of the form A∪{∗} with A∈F. Then D={(x,x):x∈X} is a subset X×Y and closedness of the projection p:X×Y→Y implies that some point (x,∗) is in its closure: D cannot be closed in X×Y, because p[D]=X is not closed in Y, as ∗∈¯p[D]=p[¯D], by closedness (and continuity) of p. This x is an adherence point of the filter, because if A∈F and x∈O where O is open in X, then O×(A∪{∗}) is basic open in X×Y and so intersects D in some (p,p), p∈X. This p∈O∩A≠∅, showing that x is an adherence point of F.

**Attribution***Source : Link , Question Author : M.Subramani , Answer Author : YuiTo Cheng*