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

Question

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.

Answer 1

Writing the composition with the helper function

$$\begin{gather} \tilde{H}\colon X\times I \to Y\times I\\ \tilde{H}(x,t) = (H(x,t),t) \end{gather}$$

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.