I am trying to give proof of some fact regarding the model structure on the category of topological spaces. I think I am just a few steps away from a solution, but i am stumped on the following problem (my topology is a bit rusty on the edges...):

Let $X$, $Y$ be topological spaces, denote by $S^{n-1}$ and $D^n$ the sphere and the disk in $\mathbb{R}^n$. Assume we have a commutative diagram as follows.

\begin{array}{c} S^{n-1} & \kern-1.5ex\xrightarrow{\ \ p\ \ }\phantom{}\kern-1.5ex & X\\ \bigg\downarrow\raise.5ex\rlap{\scriptstyle i} & & \bigg\downarrow\raise.5ex\rlap{\scriptstyle f}\\ D^n & \kern-1.5ex\xrightarrow{\ \ \alpha\ \ }\phantom{}\kern-1.5ex & Y \end{array}

Further, assume we know that $p$ is homotopic to a constant map for some homotopy $H$, which is basepoint preserving. Can we obtain maps (here denoted by question marks) giving us a diagram as follows?

\begin{array}{c} S^{n-1}\times I & \kern-1.5ex\xrightarrow{\ \ H\ \ }\phantom{}\kern-1.5ex & X\\ \bigg\downarrow\raise.5ex\rlap{\scriptstyle ?} & & \bigg\downarrow\raise.5ex\rlap{\scriptstyle f}\\ ? & \kern-1.5ex\xrightarrow{\ \ ?\ \ }\phantom{}\kern-1.5ex & Y \end{array}

I would like the vertical map to be a homotopy from $i$ to a constant path in $D^n$, or some related map, possibly keeping $\alpha$ as the horizontal map. What would/could those maps be?

EDIT: Maybe it is hard to understand what I desire without knowing what I'm trying to do.

I want to show that the map $f$ has the right lifting property relative to $i$. I know it has the right lifting property relative to the inclusions $J = \{D^n\rightarrow D^n\times I, x\mapsto (x,0)\}$. I am trying to use the homotopy $H$ to induce some map (the ones in the lower diagram) which is either in $J$ or a pushout or a transfinite composition of pushouts of elements of $J$ in order to conclude my proof.

  • $\begingroup$ By the way, does anybody have a better title for the question? $\endgroup$ Commented Apr 5, 2013 at 11:26
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    $\begingroup$ There is the trivial solution to this, taking the lower left corner to be the same as the upper right. Do you have any other requirements of the maps or of the space in the bottom left corner? $\endgroup$ Commented Apr 5, 2013 at 15:01
  • $\begingroup$ @AlešBizjak: I would like to get something related to the first diagram. Namely I would like the vertical arrow to be something like a homotopy to $D^n$ from $i$ to a constant map, or something similar. $\endgroup$ Commented Apr 5, 2013 at 15:03

1 Answer 1


I'm assuming that $H$ at time $1$ is $p$ and $H$ at time $0$ is the constant map to $x_0$ in $X$

Consider the equivalence relation $\sim$ on $S^{n-1}\times I$ given by $(q, t) \sim (q',t')$ iff $q= q'$ and $t = t'$ OR $t = t' = 0$. Note that $S^{n-1}\times I/ \sim$ is homeomorphic to $D^n$. In fact, the $q$ and $t$ are precisely polar coordinates!

Further, by the universal property of quotient maps, there is a map $\bar{H}:S^{n-1}\times I/\sim \rightarrow X$ making the obvious diagram commute.

Now, using this, replace your space $?$ by $S^{n-1}\times I/\sim$, use the quotient map for the vertical map, and the horizontal map is $f\circ \bar{H}$. The vertical map is then a homotopy from the inclusion of $S^{n-1}$ into $D^n$ to the constant map to the center of $D^n$ (after composing with a homeomorphism between $S^{n-1}\times I/\sim$ and $D^n$).

  • $\begingroup$ I'm not sure this would get me where I want to go. I would like the horizontal map to be $\alpha$ again, if possible. I'll think about it and let you know. Meanwhile +1 and I will edit my post to make clearer the context and what I desire. $\endgroup$ Commented Apr 5, 2013 at 15:17
  • $\begingroup$ I'm not sure if you can pick things differently to guarantee that that map is $\alpha$ or not. If $\alpha$ is not surjective, and one picks $H$ to homotope $p$ to a constant which is not in the image of $\alpha$, then what? $\endgroup$ Commented Apr 5, 2013 at 17:31
  • $\begingroup$ Well, actually I know that by fixing a basepoint in $S^{n-1}$, H is a basepoint preserving homotopy, thus we shouldn't have problems with that... $\endgroup$ Commented Apr 5, 2013 at 19:56

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