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When proper maps are closed

PROOF. Assuming (a) and given a proper mapf: X--Y, let Z denote the topological sum of X and Y and define g: Z-+ Y by g X =f and gi Y=iy. The g is clearly surjective and proper, hence closed, and since X is closed in Z, f is closed. Hence (a) =(b). Trivially (b) =*(c), while (c)=>(d) is immediate from the Lemma. Since (b)=*(a) is trivial it remains to deduce (b) from (d). Let f: X-Y be a proper map and F a closed subset of X. We must show f(F) is closed in Y, and since we are assuming (d), it suffices to show that if K is a compact subset of Y then Knf(F) is compact. But Knf(F) =f(f-1(K)CF).