here's the question: prove that if: f:X->Y is onto Y, closed (i.e return closed sets from given closed sets in X) and continuous then if X is Normal (satisfy axioms: T1 and T4) then also Y is Normal. Now I've showed that if X is T1 then Y is T1, but I'm having difficulty with T4. here's what I did: let F,G be disjoint closed sets of Y, then by continuity f^-1(F) and f^-1(G) are closed in X, and they are disjoint because: f^-1(FחG)=f^-1(G)חf^-1(F), now because X is T4 we have neighbourhoods of f^-1(F) and f^-1(G) which are disjoint, now I need to show that also F a G have this property, I guess I need to use the onto feature, but how? any hints?
Your neighbourhoods of f^-1(F) and f^-1(G) can be chosen to be open. Take their complements, apply f to get two closed sets in Y, then take their complements and show that these open sets separate F and G.
you mean, f^-1(F), and f^-1(G) are contained in U_G and U_F which are open and disjoint in X, and then apply the complement, and then apply f, ok that's what i did before but it got me to nowhere, i.e f(X-U_G) closed and contained in f(X-f^-1(G), and the same with F just change the G with F, then I take the complement wrt Y, but now I need to show that: G is contained in Y-f(X-f^-1(G)) (the same for F), but not sure how to do it i mean, if y is in G and not in Y-f(X-f^-1(G)) then y is in f(X-f^-1(G)) so there's x in X-f^-1(G) s.t y=f(x), but then x isnt in f^-1(G) thus f(x) isnt in G, a contradiction. ok, i see now, don't know how i got it wrong before... (-: