How to prove that a map is a Hurewicz fibration?

Question

A Hurewicz fibration is a continuous map $p:E\to B$ having the homotopy lifting property with respect to each space. Let X be a topological space. I want to prove that for all $n\geq 2$ the map $p_{n}:X^{I}\to X^{n}$ defined by $$p_{n}(\alpha)=(\alpha(0),\alpha(\dfrac{1}{n-1}),..., \alpha(\dfrac{n-2}{n-1}),\alpha(1))$$ is a hurewicz fibration. I tried to do it by induction on n (we know that $p_{2}$ is a Hurewicz fibration) but I failed. How to do it?

Answer 1

On page 50 of May's "A Concise Course in Algebraic Topology" he states in a lemma that if $A \rightarrow Y$ is a cofibration then the restriction map $B^Y \rightarrow B^A$ is a fibration.

In your case if we set $A = \{k/(n-1) \in I \mid\ 0 \leq k \leq n-1 \}$ and consider the inclusion map $A \rightarrow I$ it suffices to prove that this map is a cofibration (which is simple enough) to show that $X^I \rightarrow X^n$ is a fibration. This is because $X^A \approx X^n$ in a way that commutes with all the necessary maps proving that $X^I \rightarrow X^n$ is indeed a fibration.

By fibration and cofibration I mean Hurewicz fibration and cofibration.