# Proof of compression criterion (iff condition for representing zero in relative homotopy group)

I'm trying to prove (and understand) the compression criterion which states that a function $f\colon(I^n,\partial I^n,J^n)\to (X,A,x_0)$ represent zero in the relative homotopy group $\pi_n(X,A,x_0)$ iff it is homotopic relative $\partial I^n$ to a map $g\colon(I^n,\partial I^n,J^n)\to (X,A,x_0)$ with image contained in $A$. Note: $J^n$ here is $\partial I^n - I^{n-1}$ the same as in the book algebraic topology by Hatcher.

For the first implication: Let $g$ be such a map. Then by taking the composition of $g$ with the deformation retraction to any point that is mapped to the $x_0$ we obtain a homotopy $I^n\times I \to X$ from $g$ to the constant map. If $f$ is homotopic to $g$ relative $\partial I^n$ then $[f]=[g]$ and therefore $[f]=0$. This is how every proof I can find of this goes but it is not clear to me where exactly we use that $g(I^n)\subset A$.

The reverse implication is essentially achieved letting $H\colon I^n\times I\to X$ be a homotopy between $f$ and the constant map and then altering the domain of $H$ such that what was first the boundary of the 0-level becomes the boundary of every level while the original boundary together with the 1-level is taken to be the new 1-level. What confuses me about this part is that this modification to the domain doesn't seem to be homeomorphism.

A proof can be found here but what I wrote is based on Hatcher. https://amathew.wordpress.com/2010/10/11/the-compression-criterion/

Any clarification would be greatly appreciated.

First of all, $$J^n$$ is not $$\partial I^n-I^{n-1}$$, it is the closure $$\partial I^n-I^{n-1}$$ (in $$I^n$$), where we are identifying $$I^{n-1}$$ with $$I^{n-1} \times 0 \subset I^n$$.
I will use $$J^{n-1}$$ instead of $$J^n$$, as in Hatcher.
For the first implication, you have to use a deformation retraction $$F$$ of $$I^n$$ onto $$I^{n-1}\times 1 \subset J^{n-1}$$ (We have one obvious such deformation retraction, namely $$F:I^n \times I \to I^n, ~~(s_1,...,s_n,t)\mapsto (s_1,...,s_{n-1},(1-t)s_n+t)$$). Note that for $$[g]\in \pi_n(X,A,x_0)$$, we have by definition, $$[g]=0$$ iff $$g$$ is homotopic to the constant map $$I^n \to x_0$$ "through maps $$(I^n,\partial I^n, J^{n-1})\to(X,A,x_0)$$". Now you can check that $$[g]=0$$ via $$g \circ F$$, using the fact that $$g\circ F$$ is a homotopy through maps $$(I^n,\partial I^n, J^{n-1})\to(X,A,x_0)$$, and this is where you use that $$g$$ is a map into $$A$$.
For the reverse implication, there is no major change of the argument in Hatcher with maps of $$D^n$$. Suppose $$[f]=0$$ via a homotopy $$H:I^n\times I\to X$$ through maps $$(I^n,\partial I^n, J^{n-1})\to(X,A,x_0)$$. Then consider the composition $$H \circ G :I^n \times I \to I^n \times I \to X$$, where $$G :I^n \times I \to I^n \times I$$ is the map that sends each $$I^n \times \{ t \}$$ homeomorphically onto $$I^n \times \{ t\} \cup \partial I^n \times [0,t]$$ (think about the cube $$I^2 \times I$$ to understand this map). It is obvious that $$H\circ G$$ is a homotopy rel $$\partial I^n$$ from $$f$$ to a map into $$A$$, since $$H$$ sends $$\partial I^n \times I$$ into $$A$$ and sends $$I^n \times 1$$ to $$x_0$$.