This is from tom Dieck's book Algebraic Topology. The author uses the fact that there is a homeomorphism of pairs $k\colon (I^n\times I, (I^n\times \{0\})\cup(\partial I^n\times I))\to (I^n\times I, I^n\times \{0\})$ to prove the following theorem:
$3.2.4$. Let $p\colon E\to B$ have HLP for the cube $I^n$. If $a\colon (I^n\times \{0\})\cup(\partial I^n\times I)\to E$ is a continuous map and $h\colon I^n\times I\to B$ a homotopy such that $h\circ i = p\circ a$ where $i\colon (I^n\times \{0\})\cup(\partial I^n\times I)\to I^n\times I$ is the inclusion, then there is a homotopy $H\colon I^n\times I\to E$ with $H\circ i = a$ and $p\circ H = h$.
However, the proof is omitted and I can't guess it. The homeomorphism $k$ comes from two following theorems proved in earlier chapters:
$1$. Let $K \subseteq \mathbb{R}^n$ be a compact convex subset with nonempty interior. Then, for any $p \in \mathrm{Int}(K)$, there is a homeomorphism of pairs $(D^n, S^{n - 1})\to (K,\partial K)$ that sends $0$ to $p$.
$2$. Let $\alpha\colon D^n\times I\to I$ be given by $(x,t) = \max\{2||x||, 2 - t\}$. Then $H\colon (x,t) \mapsto (((1 + t)/\alpha(x,t))x, 2 - \alpha(x,t))$ is a homeomorphism of pairs $(D^n\times I, (D^n\times \{0\})\cup (S^{n - 1}\times I))\to (D^n\times I, D^n\times \{0\})$.
$\require{AMScd}$
The problem is to find a diagonal for the diagram \begin{CD} I^n\times 0\cup \partial I^n\times I@>a>> E\\ @V\subseteq V V @VV p V\\ I^n\times I @>h>> B. \end{CD} Now there is a homeomorphism of pairs $k:(I^{n+1},\partial I^n\times I\cup I^n\times0)\xrightarrow\cong (I^{n+1},I^n\times 0)$, with inverse, say, $l:(I^{n+1},I^n\times 0)\xrightarrow{\cong}(I^{n+1},\partial I^n\times I\cup I^n\times0)$. If we compose the above diagram with $k$ we get a commutative diagram \begin{CD} I^n\times 0@>a k|_{I^n\times0}>> E\\ @V\subseteq V V @VV p V\\ I^n\times I @>h k>> B \end{CD} which does admit a diagonal. This is simply because the square is exactly a homotopy lifting problem for $p$, and we are assuming that $p$ has the the HLP for the space $I^n$ (the definition of the HLP is at the begining of $\S$ 3.2). Let $G:I^n\times I\rightarrow E$ be a diagonal filler, i.e. $pG=hk$ and $G|_{I^n\times0}=ak|_{I^n\times0}$.
Then returning to the first diagram we consider the map $H= G l:I^n\times I\rightarrow E$. This map satisfies $$pH=pGl=hkl=h$$ and $$H|_{I^n\times 0\cup \partial I^n\times I}=(Gl)|_{I^n\times 0\cup \partial I^n\times I}=G|_{I^n\times 0}l|_{I^n\times 0\cup \partial I^n\times I}=ak|_{I^n\times0}l|_{I^n\times 0\cup \partial I^n\times I}=a.$$ Hence $H$ is a filler for the first diagram.