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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.

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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 $x_0$.

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