We say that a pair $(X,A)$ has the homotopy extension property if every pair of maps $X \times \{0\} \rightarrow Y$ and $A \times I \rightarrow Y$ that agree on $A \times \{0\}$ can be extended to a map $X \times I \rightarrow Y$.
Using this definition, I'd like to know if $(D^n,\mathring{D^n})$ has the homotopy extension property. I suspect not -- but showing this would require me to construct maps $D^n \times \{0\} \rightarrow Y$ and $\mathring{D^n} \times I \rightarrow Y$ agreeing on $\mathring{D^n} \times \{0\}$ that cannot be extended to a map $D^n \times I \rightarrow Y$.
What would such maps look like? I'm not sure what the suitable space $Y$ would be, for example.
Any help would be appreciated. Thanks!
Consider the projection $p: D^n \to D^n/\partial D^n = S^n.$ Its restriction on $\mathring{D^n}$ is a homeomorphism on $S^n\setminus\{N\}$ (for $N$ the image of $\partial D^n$), so it's contractible. Extending this contraction to the whole $D^n$ would mean showing that $p$ is null-homotopic. But it's not: it generates $\pi_n(S^n, N) \simeq \mathbb{Z}.$ Intuitively, if you move the inner part of the disk's image around then the boundary may not be able to follow its trajectory because it's thin (closed, not open) and can accordingly be torn apart. E.g. when you contract $S^n\setminus\{\text{North pole}\}$ to the south pole, the boundary, which is mapped to $\{N\}$, can't decide in which direction to go toward the south pole.Let $(X,A) := (D^n, \mathring{D^n})$ and consider the "hat" $Y:= A \times I \cup X \times \{0\}.$ There is a map $X \times \{0\} \hookrightarrow Y$ and a constant homotopy $A \times I \hookrightarrow Y.$ Extending this would give a map $X \times \{1\} \to Y$ such its restriction on $A \times \{1\}$ maps it onto the top of the hat. Compose this with $Y \hookrightarrow X \times I.$ By continuity the resulting map $X \times \{1\} \to X \times I$ has no choice but to be the inclusion in the $1$-th layer ("top"). This contradicts the construction by which it also factors through $Y$.