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.
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 $D^n$).