Recall the Compression Criterion:
A map $f\colon [D^n, S^{n-1},s_0] \rightarrow [X, A,x_0]$ represents zero in $\pi_n(X, A,x_0)$ if and only if $f$ is homotopic relative $S^{n-1}$ to a map $g\colon D^n \rightarrow A$.
It is a basic tool in basic Homotopy Theory, and therefore my intention is to fully understand the mechanics behind it. So I started looking at the proof carefully highlighting where does each hypothesis come into play and the following question arose immediately:
Why do we need that $$f\sim_{\text{rel}S^{n-1}} g$$ instead of simply $$f\sim g$$
Obviously by $f\sim g$ I mean an homotopy of triple $[D^n,S^{n-1},s_0]\times I \to [X,A,x_0]$ which by definition is a a continuous family of functions of triples of spaces. BUT it's less than asking that the homotopy fixes point-wise $S^{n-1}$.
The only direction of the proof which (maybe) has to be changed is clearly the "easy" one, because in the other direction (starting with a function homotopic to the constant one and building another homotopic function whose image is contained in $A$) we are actually prove more than what is needed (exhibiting a homotopy relative to $S^{-1}$, instead of a homotopy of triple).
So let $f\sim g$ via a triple homotopy (this is the definition of representing the same element in $\pi_n(X,A,x_0)$ as written in here for example). Now the same argument can be applied to $g$ to retract it to a point, and now gluing the homotopy from $f$ to $g$ and from $g$ to $x_0$ proves the claim.
I don't see where does the hypothesis about the homotopy comes into play. Could someone clarify this detail to me?