I have a question in a step of a proof in P. May's "A Concise Course in Algebraic Topology" (page 70):
Why does the condition (iii) imply that if $e \circ g $ null homotopic then also $g$?
Well, (iii) imply that $g$ factorieses over $CS^n$ (the "mapping cone") by $\tilde{g}$, but how does it induce null homotopy?
Suppose you have a map $g:X\to Y$ which extends to a map $\tilde{g}:CX\to Y$ on the cone $CX=X\times I/X\times\{1\}$ (identifying $X$ with $X\times\{0\}$ in $CX$). Composing $\tilde{g}$ with the quotient map $p:X\times I\to CX$, we get a map $H:X\times I\to Y$. Since $\tilde{g}$ is an extension of $g$, $H(x,0)=g(x)$ for each $x\in X$. Also, since $p$ is constant on $X\times\{1\}$, so is $H$. So $H$ is a homotopy from $g$ to a constant map.