Projection map being a closed map

Let π:X×YX 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.


Suppose ZX×Y is closed, and suppose x0Xπ[Z].
For any yY,(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)=xU. Then yV(yi) for some i, and as xUU(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 Uπ[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 AF. Then D={(x,x):xX} is a subset X×Y and closedness of the projection p:X×YY 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 AF and xO where O is open in X, then O×(A{}) is basic open in X×Y and so intersects D in some (p,p), pX. This pOA, showing that x is an adherence point of F.

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

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