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.

  • 1
    $\begingroup$ Write $\tilde{H}\colon X\times I \to Y\times I, \tilde{H}(x,t) = (H(x,t),t)$. $\endgroup$ Jul 25, 2014 at 19:41
  • $\begingroup$ @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. $\endgroup$ Jul 25, 2014 at 19:43
  • $\begingroup$ $K^{-1}(U)$ need not be a product of open sets, it might be an arbitrary union of such products. $\endgroup$ Jul 25, 2014 at 19:44
  • $\begingroup$ @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. $\endgroup$ Jul 25, 2014 at 19:48
  • $\begingroup$ 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. $\endgroup$ Jul 25, 2014 at 22:16

1 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.


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