How to show this function defined using a homotopy is continuous?

Question

Suppose there exists a homotopy $H$ between continuous functions $f,g:X\longrightarrow Y$, where $X,Y$ are non-empty topological spaces. Consider the function $h:X\times [-1,1]\times [0,1]\longrightarrow Y\times[-1,1]$ defined by $h(x,s,t)=(H(x,t),s)$. I want to show that this function is continuous, but I am not sure how to.

I tried to show that such a map is "induced" (and is therefore automatically continuous) by the continuous map $f':X\times [-1,1]\longrightarrow Y\times [-1,1]$ given by $f'(x,t)=(f(x),t)$, under some identification map from $X\times [-1,1]$ to $X\times [-1,1]\times [0,1]$, but I am not sure if such an identification map exists.

Or is there a simpler argument along the lines of "$h$ is continuous because it is defined using $H$, which is continuous because its a homotopy"? I feel this is not sufficient mainly due to the fact that we 'swap' the order of arguments $s$ and $t$ in some way.

Can anyone provide some clarity on this problem? Thanks

Answer 1

You only need to know the following fact about the product topology:

A map $\phi : A \to P = \prod_{j \in J} X_j$ is continuous if and only if all component functions $p_j \circ \phi : A \to X_j$ are continuous. Here, $p_j : P \to X_j$ denotes projection onto the $j$-th component.

Note that this statement includes the continiuity of the $p_j$. Simply take $\phi =id_P$ which is clearly continuous. In fact, the product topology is introduced as the coarsest topology on $P$ making all $p_j$ continuous.

Thus your function $h$ is continuous iff $h_1(x,s,t) = H(x,t)$ and $h_2(x,s,t) = s$ are continuous. But $h_2$ is the projection onto the second coordinate, hence continuous. Moreover, $h_1 = H \circ p$, where $p : X\times [-1,1]\times [0,1] \to X\times [0,1], p(x,s,t) = (x,t)$. But $p$ is continuous by the above statement. Thus $h_1$ is continuous.