[proofplan] We regard each finite poset as a category with one morphism $x\to y$ exactly when $x\le y$. Under this identification, the order complex is the nerve of the corresponding poset category, and an order-preserving map is a functor. The fiber hypothesis is exactly the lower-comma hypothesis in the finite-poset form of Quillen's Theorem A, so that theorem applies and gives that the induced map of nerves, equivalently the induced map of order complexes, is a homotopy equivalence. [/proofplan] [step:Interpret the poset map as a functor between poset categories] Let $\mathcal P$ denote the category associated to $P$: its objects are the elements of $P$, and for $p_0,p_1\in P$ there is a unique morphism $p_0\to p_1$ if and only if $p_0\le_P p_1$. Let $\mathcal Q$ denote the analogous category associated to $Q$. Because $f:P\to Q$ is order-preserving, it determines a functor \begin{align*} F:\mathcal P\to\mathcal Q \end{align*} defined on objects by $F(p)=f(p)$ and on the unique morphism $p_0\to p_1$ by the unique morphism $f(p_0)\to f(p_1)$. This is well-defined since $p_0\le_P p_1$ implies $f(p_0)\le_Q f(p_1)$. The nerve $N(\mathcal P)$ is the simplicial set whose nondegenerate simplices are chains in $P$, hence its geometric realization is the order complex $\Delta(P)$ as defined in the theorem statement. Similarly, the geometric realization of $N(\mathcal Q)$ is $\Delta(Q)$. Therefore the functor $F$ induces precisely the continuous simplicial map \begin{align*} \Delta(f):\Delta(P)\to\Delta(Q). \end{align*} [guided] A finite poset can be viewed as a category in a canonical way: elements are objects, and the order relation supplies the morphisms. Thus for $P$ we define a category $\mathcal P$ whose object set is $P$, and where there is exactly one morphism $p_0\to p_1$ precisely when $p_0\le_P p_1$. The category $\mathcal Q$ is defined from $Q$ in the same way. The order-preserving map \begin{align*} f:P\to Q \end{align*} then becomes a functor \begin{align*} F:\mathcal P\to\mathcal Q. \end{align*} On objects we set $F(p)=f(p)$. If there is a morphism $p_0\to p_1$ in $\mathcal P$, then $p_0\le_P p_1$. Since $f$ is order-preserving, $f(p_0)\le_Q f(p_1)$, so there is a unique morphism $f(p_0)\to f(p_1)$ in $\mathcal Q$. This defines $F$ on morphisms and proves that the functor is well-defined. The reason for passing to categories is that Quillen's Theorem A is naturally stated for functors and nerves. The nerve $N(\mathcal P)$ records chains in $P$: a nondegenerate $k$-simplex is a strict chain \begin{align*} p_0<_P p_1<_P \cdots <_P p_k. \end{align*} Thus the geometric realization of $N(\mathcal P)$ is exactly the order complex $\Delta(P)$. The same identification gives $|N(\mathcal Q)|=\Delta(Q)$. Under these identifications, the map induced by the functor $F$ is the continuous map on order complexes induced by $f$. [/guided] [/step] [step:Identify the lower comma categories with the stated fibers] Fix $q\in Q$. The lower comma category $F\downarrow q$ has objects pairs $(p,\alpha)$ where $p\in P$ and $\alpha:F(p)\to q$ is a morphism in $\mathcal Q$. Since $\mathcal Q$ is the category associated to the poset $Q$, such a morphism exists if and only if \begin{align*} f(p)\le_Q q. \end{align*} It is unique when it exists. Hence the object set of $F\downarrow q$ is canonically identified with \begin{align*} f^{-1}(Q_{\le q})=\{p\in P:f(p)\le_Q q\}. \end{align*} Under this identification, the morphisms are exactly the order relations inherited from $P$. Therefore the nerve of $F\downarrow q$ has geometric realization \begin{align*} |N(F\downarrow q)|=\Delta(f^{-1}(Q_{\le q})). \end{align*} By hypothesis, this space is contractible for every $q\in Q$. [/step] [step:Apply Quillen's Theorem A for finite posets] We use Quillen's Theorem A in its finite-poset form: if $F:\mathcal P\to\mathcal Q$ is a functor between finite poset categories and, for every object $q\in\mathcal Q$, the nerve of the lower comma category $F\downarrow q$ has contractible geometric realization, then the induced map \begin{align*} |N(F)|:|N(\mathcal P)|\to |N(\mathcal Q)| \end{align*} is a homotopy equivalence. The hypotheses of this result are satisfied: $\mathcal P$ and $\mathcal Q$ are finite poset categories, $F:\mathcal P\to\mathcal Q$ is the functor induced by the order-preserving map $f$, and the previous step identifies each lower comma realization $|N(F\downarrow q)|$ with the contractible order complex $\Delta(f^{-1}(Q_{\le q}))$. Hence \begin{align*} |N(F)|:|N(\mathcal P)|\to |N(\mathcal Q)| \end{align*} is a homotopy equivalence. [guided] Quillen's Theorem A is the mechanism that converts contractibility of all lower fibers into a homotopy equivalence on nerves. In the finite-poset form used here, its input is a functor \begin{align*} F:\mathcal P\to\mathcal Q \end{align*} between finite poset categories such that, for every object $q\in\mathcal Q$, the geometric realization $|N(F\downarrow q)|$ is contractible. Its conclusion is that the induced continuous map on geometric realizations of nerves, \begin{align*} |N(F)|:|N(\mathcal P)|\to |N(\mathcal Q)|, \end{align*} is a homotopy equivalence. We verify the hypotheses one by one. The categories $\mathcal P$ and $\mathcal Q$ are finite because their object sets are the finite sets $P$ and $Q$, and each hom-set has either one morphism or no morphisms. The map $F:\mathcal P\to\mathcal Q$ is a functor because it was constructed from the order-preserving map $f:P\to Q$. Finally, for each $q\in Q$, the previous step identifies the lower comma category $F\downarrow q$ with the subposet $f^{-1}(Q_{\le q})$, so its nerve has geometric realization \begin{align*} |N(F\downarrow q)|=\Delta(f^{-1}(Q_{\le q})). \end{align*} The theorem hypothesis says that this space is contractible for every $q\in Q$. Therefore all hypotheses of Quillen's Theorem A are satisfied, and the conclusion gives that \begin{align*} |N(F)|:|N(\mathcal P)|\to |N(\mathcal Q)| \end{align*} is a homotopy equivalence. [/guided] [/step] [step:Translate the conclusion back to order complexes] Using the identifications \begin{align*} |N(\mathcal P)|=\Delta(P) \end{align*} and \begin{align*} |N(\mathcal Q)|=\Delta(Q), \end{align*} the homotopy equivalence $|N(F)|$ is precisely the induced map \begin{align*} \Delta(f):\Delta(P)\to\Delta(Q). \end{align*} Therefore $\Delta(f)$ is a homotopy equivalence, as required. [/step]