Lemma 4.22. Let $X$ be a space and, for $i=0,1$, let $\lambda _i:X\rightarrow X\times I$ be defined by $x\mapsto (x,i)$. If $H_n (\lambda _0)=H_n(\lambda _1)$, then $H_n(f)=H_n(g)$ whenever $f,g:X\rightarrow Y$ are homotopic.
What is the geometric interpretation of the equality $H_n (\lambda _0)=H_n(\lambda _1)$?
I obtained nice intuition from Chapter I, section 2 of Kamps & Porter's Abstract Homotopy and Simple Homotopy Theory. There, the basic theory of cylinder-based homotopy is developed. $\lambda _0,\lambda _1$ are natural maps that "embed" $X$ into the cylinder $X\times I$. Part of the idea of homotopy $h:f\simeq$ is that one can recover the functions $f,g$ (top and bottom bases of the cylinder) by precomposing the homotopy with $\lambda_0 ,\lambda _1$. Now the punchline is the functoriality of homology. Rewrite $f,g$ as the precompositions of the homotopy with $\lambda _i$ - it's now obvious that if you cannot tell apart $\lambda _0 ,\lambda _1$ in homology, there's no hope of telling apart $f,g$, which is exactly the point of the lemma.