I have the following problem:

I have $\beta:D^n\to Z$ such that $\beta\circ i\simeq h$, where $i:S^{n-1}\hookrightarrow D^n$ is the inclusion and $h:S^{n-1}\to Z$ some map. Then is there a $\beta':D^n\to Z$ with $\beta\simeq\beta'$ such that $\beta'\circ i=h$?

I've looked for the answer a bit online, and I've understood that this condition is being a cofibration, but I didn't explicitly see the proof that $i:S^{n-1}\hookrightarrow D^n$ is a cofibration.


1 Answer 1


It clearly suffices to show that there exists a retraction $r : D^n \times I \to D^n \times \{0\} \cup S^{n-1} \times I$. In fact, the map $\beta$ and the homotopy $\varphi: \beta \circ i \simeq h$ can be pasted to a continuous $\phi : D^n \times \{0\} \cup S^{n-1} \times I \to Z$ and $\phi \circ r$ is what you desire.

Define $$r(x,t) = \begin{cases} (\frac{2x}{2-t},0) & \lVert x \rVert \le \frac{2-t}{2} \\ (\frac{x}{\lVert x \rVert},2 - \frac{2-t}{\lVert x \rVert}) & \lVert x \rVert \ge \frac{2-t}{2} \end{cases} $$ This is a well-defined continuous map because

  1. $\lVert \frac{2x}{2-t} \rVert \le 1$ for $\lVert x \rVert \le \frac{2-t}{2}$

  2. $1 \ge t \ge 2 - \frac{2-t}{\lVert x \rVert} \ge 0$ for $1 \ge \lVert x \rVert \ge \frac{2-t}{2}$

  3. For $\lVert x \rVert = \frac{2-t}{2}$ both parts yield the same value.

For $t=0$ we get $\frac{2-t}{2} = 1$ and thus $r(x,0) = (x,0)$ for all $x \in D^n$. For $x \in S^n$ we have $r(x,t) = (x,t)$. Thus $r$ is a retraction.

There is a nice geometric interpretation of $r$: For each $(x,t)$ consider the line $L_{x,t}$ through $(0,2)$ and $(x,t)$. This line intersects $D^n \times \{0\} \cup S^{n-1} \times I$ in a single point $r(x,t)$.


It is very easy to see that for a closed subspace $A \subset X$ the inclusion $i : A \to X$ is a cofibration if and only if $X \times \{0\} \cup A \times I$ is a retract of $X \times I$. See for example Retraction induces Cofibration . This is also true for non-closed $A \subset X$, but the proof is sophisticated. See

Strøm, Arne. "Note on cofibrations II." Mathematica Scandinavica 22.1 (1969): 130-142.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.