Question on homotopic maps

algebraic-topologyhomotopy-theory

I was going through some introductory algebraic topology. The book I am reading says that:

If there is a homotopy $H: X\times I \rightarrow Y$ where $I = [0,1]$, then for each $t \in I$, there is a continuous map $h_t: X\rightarrow Y$ defined by

$$h_t(x) = H(x,t) \text{for all } x \in X$$

Now I was wondering how to show that this $h_t$ is continuous.

I know that $H$ is continuous by the definition of homotopy.

To prove $h_t$ is continuous, I will have to show that for each open set $U$ in $Y$, $h_t^{-1}(U)$ is open in $X$.

How can I show that or is there any other method?

Please, help

Best Answer

You can write the function $h_t : X \to Y$ as a composition of two continuous functions, and then apply the theorem that a composition of continuous function is continuous.

  1. First, the function $X \to X \times [0,1]$ defined by $x \mapsto (x,t)$;
  2. followed by the function $H : X \times [0,1] \to Y$.

Continuity of the second function $H$ is given to you. Continuity of the first function can, if you like, be easily proven using the open set definition, combined with the definition of the product topology on $X \times [0,1]$.

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