Does $(D^n,S^{n-1})$ have the homotopic extension property? Recall that we say that $(X,A)$ has homotopy extension property if  $X\times I$  retracts to $X\times\{0\}\cup A\times I$ . Is it true that 

$(D^n,S^{n-1})$ has homotopic extension property? 

I'm trying to prove this just using definitions but i am unable to do this.Any ideas?
 A: Here's a direct proof. The intuition is that we can "flatten out" the interior and send everything else to the boundary: Imagine taking a can of play-doh and crushing it to the bottom and sides. So embed $D^n\times I$ into $\mathbb{R}^{n+1}$ by mapping $(p,t)\mapsto (p,t/2)$ and then use stereographic projection to smoosh the interior of $D^n\times I$ to the "bottom" and "sides" of the can.
A: Here is a picture for $n=1$. (Please excuse the hand drawing.) In general, you can project any point in $D^n \times I$ onto $D^n \times 0 \cup S^{n-1} \times I$ along a ray emanating from the point $(0,2)$.

You could write a formula for the retraction, but it is more complicated than the picture:
$$ r \colon D^n \times I \to D^n \times 0 \cup S^{n-1} \times I$$
$$ r(x,t) = \begin{cases}
\Bigl(\frac{2x}{2-t},0 \Bigr) & |x| \le \frac{2-t}{2} \\
\Bigl(\frac{x}{|x|},2-\frac{2-t}{|x|}\Bigr) & |x| \ge \frac{2-t}{2}.
\end{cases}$$
A: Yes it does!
First of all, let's state the original definition of the homotopy extension property (which explains its name). We say that $(X,A)$ has the homotopy extension property if for any space $Y$ and any homotopy $f_t\colon A\to Y$, given a map $\tilde{f_0}\colon X\to Y$ such that $f_0 = \left.\tilde{f_0}\right|_A$, there exists a homotopy $\tilde{f}_t\colon X\to Y$ such that $f_t = \left.\widetilde{f}_t\right|_A$.
Now assume you have a homotopy $f_t\colon S^{n-1}\to Y$ and some map $\tilde{f}_0\colon D^n \to Y$ such that $f_0 = \left.\tilde{f}_0\right|_{S^{n-1}}$.
It is convenient to consider the disk of radius $2$:
$$D_2^n = \{ \mathbf{x} \in \mathbb{R}^n \mid \|\mathbf{x}\| \le 2 \},$$
and a map $D_2^n \to Y$
$$h (\mathbf{x}) = \begin{cases}
\tilde{f}_0 (\mathbf{x}), & \|\mathbf{x}\| \le 1,\\
f_{\|\mathbf{x}\| - 1} (\mathbf{x} / \|\mathbf{x}\|), & 1 \le \|\mathbf{x}\| \le 2\\
\end{cases}$$
(check that it is continuous and well-defined).
Now we can take
$$\tilde{f}_t = h ((1+t)\,\mathbf{x}).$$
For $t = 0$ it is indeed $\tilde{f}_0$, and on the sphere (when $\|\mathbf{x}\| = 1$) they coincide: $f_t = \left.\tilde{f}_t\right|_{S^{n-1}}$.

P.S. The homotopy extension property for $(D^n, S^{n-1})$ is used to show that any CW pair has it.
