Talk:Open and closed maps

Sufficient for continuity?

Regarding "...a function f : X → Y is continuous...if the preimage of every closed set of Y is closed in X." at the end of the second paragraph:

Forgive me if I'm mistaken, but in general, preimages of closed sets being closed does not ensure continuity. E.g. for f: [0,1] -> (0,1] where f(x) = x (for x != 0) and f(0) = 1/2, f is not continuous. Perhaps something is assumed that I missed? 203.150.100.189 08:38, 20 March 2007 (UTC)[reply]

Please read again: "if the preimage of every open set of Y is open in X". For your example, the preimage of ]1/4,3/4[ is then {0}U]1/4,3/4[ which is not open.--133.11.80.84 07:05, 3 September 2007 (UTC)[reply]

203: Pass to complements. Suppose f is continuous and take a closed set V. Then the complement of V, which I'll denote C(V), is open, so it pulls back to an open set. But the preimage of V is the complement of the preimage of C(V), so it's closed. The other way: Suppose f pulls back closed sets to closed sets. A similar argument gives you that f pulls back open sets to open sets, so f is continuous. HTH.

Druiffic (talk) 04:26, 15 November 2008 (UTC)Druiffic[reply]

Unless I'm going crazy, "An open map is also closed if and only if it is surjective" is definitely not true. Just include X into two (disjoint) copies of itself. —Preceding unsigned comment added by 24.19.0.156 (talk) 00:54, 8 February 2011 (UTC)[reply]