A follow up to "Showing that a sequence of a cofibration is exact" After reading through the explanation that Tyrone gave on this page Showing that a sequence of a cofibration is exact.
I am curious on how one would use that information to fully prove that if $A \rightarrow B$ is a cofibration with cofiber $C$, then for any pointed space $X,$ the sequence $[C,X] \rightarrow [B,X] \rightarrow [A,X]$ is an exact sequence of pointed sets.
 A: Recall that $C_f$ is obtained from $B$ and $CA:=A\times[0,1]/A\times\{1\}$ by identifying $(a,0)\in CA$ with $f(a)\in B$. Next, recall that Tyrone defines a cofiber sequence as follows

A cofiber sequence is a sequence $A\xrightarrow fB\xrightarrow gC$, together with an homotopy $H:gf\rightarrow *$ such that the induced map $\chi:C_f\rightarrow C$ is an homotopy equivalence.

In particular, since we're looking at homotopy classes of maps, precomposition with $\chi$ induces an isomorphism $$[C,X]\xrightarrow\simeq [C_f,X]$$ and the sequence $[C,X]\rightarrow [B,X]\rightarrow [A,X]$ is equivalent to $[C_f,X]\rightarrow[B,X]\xrightarrow{f_*}[A,X]$, so we only need to show exactness of the latter to derive exactness of the former.
It should be clear now how you conclude exactness from this property of $C_f$: Since $C_f$ is obtained by gluing $B$ and $CA$, a morphism $C_f\rightarrow X$ is equivalent to morphisms $\beta:B\rightarrow X$ and $\alpha:CA\rightarrow X$ that are compatible with the way $B$ and $CA$ are glued, meaning $$\beta(f(a))=\alpha(a,0)$$Moreover, since $CA$ is obtained by taking $A\times I$ and collapsing $A\times\{1\}$ to a point, a morphism $\alpha:CA\rightarrow X$ is equivalent to a morphism $H:A\times I\rightarrow X$ such that $H(-,1)=*$ (in this general scenario, $*$ need not be the point of $X$, but in the proof of exactness it will be), hence an homotopy between $*$ and $H(-,0)=\alpha(-,0)=\beta\circ f$. Do you see how to conclude?
