# Show that composition of continuous function is continuous in product topology.

Suppose $H: X \times I \to Y$ is a continuous map of topological spaces $X,Y$ and $I = [0,1]$. And suppose $K: Y \times I \to Z$ is also a continuous map of topological spaces.

I want to show that the composition $L(x,t) = K(H(x,t),t)$, $L: X \times I \to Z$, is continuous. I am unsure of how to think about composing functions in the "input".

Attempt: my approach is the show that for an open set $U \subseteq Z$, the preimage $L^{-1}(U)$ is open. $K$ is continuous, so $K^{-1}(U)$ is open.

Now, $K^{-1}(U) \subseteq Y \times I$, so it looks like a product of open sets. i.e. $K^{-1}(U) = V \times T$. But I am not sure how to proceed from here. If this was a map into product $f: X_1 \to X_2 \times X_3$ then I know I can simply check the component functions. The situation here is like a map from a product instead.

• Write $\tilde{H}\colon X\times I \to Y\times I, \tilde{H}(x,t) = (H(x,t),t)$. Jul 25, 2014 at 19:41
• @DanielFischer - Ah thanks! I totally forgot we could do that. Its one line but if you post is as an answer I will accept and +1. Jul 25, 2014 at 19:43
• $K^{-1}(U)$ need not be a product of open sets, it might be an arbitrary union of such products. Jul 25, 2014 at 19:44
• @BenMillwood - Thanks for pointing that out! (finite) Product Topology is generated by product of open sets, which means it could be arbitrary union or finite intersection etc of such products. Jul 25, 2014 at 19:48
• The finite intersections are fine though, because the intersection of two of those sets is another one of those sets, and an intersection of unions is a union of intersections. Jul 25, 2014 at 22:16

$$\begin{gather} \tilde{H}\colon X\times I \to Y\times I\\ \tilde{H}(x,t) = (H(x,t),t) \end{gather}$$
makes the continuity of $L$ immediate. $L = K\circ \tilde{H}$, and both components of $\tilde{H}$ are directly seen to be continuous, so $\tilde{H}$ is continuous.