Lemma 1. Let X, Y be topological spaces; let . Then the following are equivalent.

(i) f is continuous.

(ii) For each and each neighborhood V of , there is a neighborhood U of x such that .

Let W be a neighborhood of for any and a fixed point . Since F is continuous, there exists a neighborhood V of such that by lemma 1. Since is an open set containing in , there exists an open set U in X containing x such that , where is an open projection map.

Now, for any neighborhood W of , we have an open set containing x such that .

Thus, h is a continous by lemma 1.

It is pretty similar to show that k is continuous.