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 '14 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 '14 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 '14 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 '14 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 '14 at 22:16

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.