In Quillen's original 1978 paper, he proved the following proposition:
Let $f: X \to Y$ be a poset map (this means that the map is order preserving; that is, $x_1 \leq x_2$ implies $f(x_1)\leq f(x_2)$). Let $Y_{\leq y} = \{q \in Y: q \leq y\}$. Assume $f^{-1}(Y_{\leq y})$ is contractible for all $y \in Y$. Then $f$ is a homotopy equivalence.
A corollary of this proposition is the following: Let $f: X \to X$ be a poset map. If $f(x) \leq x$ for all $x \in X$, then $X$ deformation retracts to $f(X)$.
When we talk about "homotopy equivalence" or "deformation retract" here, we mean the geometric realization of the poset. For a poset $X$, the vertices of its geometric realization are the elements of $X$ and the faces are the finite chains. I think in the corollary, I can prove that the fibers are all contractible, which shows that $X$ and $f(X)$ are homotopy equivalent, but I don't see why $f$ is a deformation retract.
Corollary 0.20 in Hatcher says that for CW complexes, if the inclusion of a subcomplex is a homotopy equivalence, then the subcomplex is a deformation retract.