I have this question from a Quillen's paper Homotopy Properties of the Poset of Nontrivial p-Subgroups of a Group:
Homotopy property: If $f,g\colon X\rightarrow Y$ are maps of posets such that $f(x)<g(x)$ for all $x\in X$, then $\mid f\mid $ and $\mid g\mid$ are homotopic.
The proof states that this comes from the identity between geometric realisations of posets $$ \mid X \times Y\mid\simeq \mid X\mid \times \mid Y\mid $$
And considering a poset $\{0<1\}$, the functions $f,g$ determine a map
$$ \{0<1\} \times X\rightarrow \times Y$$
But I don't understand why this is a proof of this. Could anyone explain me the proof of this property?
Let $I = \{ 0 < 1 \}$ be the poset. Put the product order on $I \times X$, i.e. $(s,x) \le (t,y) \iff s \le t \land x \le y$. Then the data of $f,g: X \to Y$ such that $f(x) < g(x)$ induces a poset map $h : I \times X \to Y$ given by $h(0,x) = f(x)$ and $h(1,x) = g(x)$.
Let $F = |f|$ and $G = |g|$. If you take the realization of $h$, you get a map $$H = |h| : |I \times X| \to |Y|.$$ You know that $|I \times X| \cong |I| \times |X|$. Moreover it's pretty clear that $|I| \cong [0,1]$, the interval. So your map $H$ is in fact a map $[0,1] \times |X| \to |Y|$. It isn't hard to check that $H(0,x) = F(x)$ and $H(1,x) = G(x)$ for $x \in |X|$ (and where $0,1 \in [0,1]$, not $I$, it's a little bit of yoga). So you do get a homotopy.