I have a solution, but i'm not sure whether it is correct:
We want to show f is proper <=> f is closed.
Let V be a closed subset of X.
f(V) is closed in Y <=> Y-f(V) is open in Y.
so let y Y-f(V) a arbitrary point. We know Y is Hausdorff, hence for every point v f(V) we have a open set of v and a open set U of y, s.t.
(*)hence it exist a nbh U of y , s.t. U f(V) =
hence f(V) is closed.
I'm not sure whether this prove is correct.
I mean, i havnÄt use locally compactness of Y. So i think this can't be correct.
the part with (*) above isn't correct, or? how can i involve locally compactness in that point, such that my proof is correct?