I'm trying to prove that every map $f:X\to Y$ is homotopic to fibration (that for every $f:X\to Y$ exists $X'$ and $f':X'\to Y$ such that $f'$ is Hurewitz fibration and $f'H$ is homotopic to $f$).
Let $X'=\{(x,w)\in X\times Y^I:f(x)=w(0)\}$, $H(x)=(x,f(x))$, $H^{-1}(x,w)=x$ and $f'(x,w)=w(1)$. I've checked that $H$ and $H^{-1}$ are homotopy inverse maps and that $f'H=f$.
It remains to prove that $f'$ is fibration. How to check that $f'$ has the HLP?
Any help is welcome. Thanks in advance.
You can directly find a solution to the homotopy lifting problem. Suppose you have a commutative diagram $$\begin{array}{ccc} A\times\{0\} & \stackrel{g}{\longrightarrow}& X'\\ \downarrow{i} & & \downarrow{f'}\\ A\times I & \stackrel{H}{\longrightarrow} & Y \end{array}$$ where $i$ is the inclusion to the base of the cylinder $A\times I$. The map $g$ has two components: $$ g(a,0)=(g_1(a),g_2(a))\in X'\subseteq X\times Y^I, $$ and, since $g(a,0)\in X'$ we must have $g_2(a)(1)=f(g_1(a))$. By commutativity of the diagram we also know that $$ H(a,0)=f'(g(a,0))=g_2(a)(1) $$ Hence the endpoint of the path $g_2(a)$ is the starting point of the path $H(a,-)$. After this observation you can define the lift $\tilde{H}:A\times I\rightarrow X'$. Notice that $\tilde{H}$ has two components $$ \tilde{H}_1:A\times I\rightarrow X\\ \tilde{H}_2:A\times I\rightarrow Y^I $$ Define $$ \tilde{H}_1(a,t):=g_1(a) $$ and $\tilde{H}_2(a,t)$ as the path which runs along the path $g_2(a)$ and then continues along $H(a,-)$, ending at $H(a,t)$ (remember, the endpoint of $g_2(a)$ is the beginning of $H(a,-)$). Explicitly: $$ \tilde{H}_2(a,t)(s):= \begin{cases} g_2(a)((1+t)s) & \text{ if } 0\leq s\leq 1/(1+t)\\ H(a,((1+t)s-1) & \text{ if } 1/(1+t)\leq s\leq 1. \end{cases} $$ I will leave to you to verify that $\tilde{H}$ is indeed a lift and that it is continuous.