Property of Homology: group isomorphism

Question

I have this proposition, but I don't understand how they use the axiom 5, since in the axiom 5; $f,g: (X,A)\rightarrow (Y,B)$ and in the theorem we have $f:(X,A)\rightarrow (Y,B)$, $g:(Y,B)\rightarrow (X,A)$?

And the axiom 5 is:

Answer 1

The $f$ and $g$ appearing in the proof are not the same as the $f$ and $g$ appearing in the axiom. We have $g\circ f\sim\mbox{Id}_{X,A}\colon (X,A)\to (X,A)$ and so axiom $5$ implies that $(g\circ f)_*=(\mbox{Id}_{X,A})_*$, the other axioms then give us that $(g\circ f)_*=g_*\circ f_*$ by funtoriality, and $(\mbox{Id}_{X,A})_*=\mbox{Id}_{H_k(X,A)}$ also by functoriality. Similarly for $f\circ g$. The proof is completed by using the fact that an isomorphism is defined to be an arrow which has another arrow acting as a right and left inverse, in this case $g_*$ is the right and left inverse of $f_*$.