Let $f:S^n\to S^n$ be an arbitrary continuous map. It is a part of Exercise 15 in Hatcher's AT, chapter 4.1, to show that $f$ is homotopic to a map $g$ such that there exists a point $q\in S^n$ with $g^{-1}(q) = \{p_1, \dots ,p_k\}$ and $g$ is an invertible linear map near each $p_i$. Hatcher suggests to use one of the following two theorems:
Theorem 1 (Simplicial Approxmiation). If $K$ is a finite simplicial complex and $L$ is an arbitrary simplicial complex, then any map $f :K\to L$ is homotopic to a map that is simplicial with respect
to some iterated barycentric subdivision of $K$.
Theorem 2. Let $f : I^n\to Z$ be a map, where $Z$ is obtained from a subspace $W$ by attaching a cell $e^k$. Then there is a homotopy $f_t:(I^n, f^{-1}(e^k))\to (Z, e^k)$ rel $f^{-1}(W)$ from $f = f_0$ to a map $f_1$ for which there is a polyhedron $K \subset I^n$ such that:
(a) $f_1(K)\subset e^k$ and $f_1|_K$ is PL with respect to some identification of $e^k$ with $\Bbb R^k$.
(b) $K\supset f^{-1}(U)$ for some nonempty open set $U$ in $e^k$.
Here are the definitions of polyhedron and PL map in Hatcher's book: Define a polyhedron in $\Bbb R^n$ to be a subspace that is the union of finitely many convex polyhedra, each of which is a compact set
obtained by intersecting finitely many half-spaces defined by linear inequalities of the form $\sum_i a_ix_i \leq b$. By a PL (piecewise linear) map from a polyhedron to $\Bbb R^k$ we shall mean a map which is linear when restricted to each convex polyhedron in some such decomposition of the polyhedron into convex polyhedra.
I can see that both theorems are applicable in the situation above, but how can we choose a point $q$ satisfying the required properties?
(Actually using differential topology I can show that the result is true: every continuous map $f:S^n\to S^n$ is homotopic to a smooth map $g:S^n\to S^n$; just take $q$ to be any regular value of $g$.)
Best Answer
Via simplicial approximation you can find a map $g$ which is homotopic to $f$ and is simplicial with respect to some iterated barycentric subdivision of $K$. Denote this iterated barycentric subdivision by $K^*$.
If $g$ is not surjective, then pick $q \notin g(S^n)$. Then $g^{-1}(q) = \emptyset$ which is the trivial case with $0$ preimage points. Note that any non-surjective $g$ represents $0 \in \pi_n(S^n)$ which is a trivial multiple of the identity map.
If $g$ is surjective, then pick any $n$-simplex $\Delta$ of $K$ and let $q$ be an interior point of $\Delta$. Since $g$ is simplicial, it maps each simplex of $L$ linearly onto a simplex of $K$. The set of simplices $\Delta^*$ of $K^*$ such that $q \in g(\Delta^*)$ is finite ($K^*$ is a finite simplicial complex) and non-empty ($S^n$ is the union of all images of simplices in $K^*$). Let $\Delta_1^*,\ldots, \Delta_k^*$ be these finitely many simplices ($k \ge 1$). No $\Delta_i^*$ can have dimension $< n$ because in that case $g(\Delta_i^*)$ would be a simplex of dimension $< n$ which cannot contain $q$. Thus the $\Delta_i^*$ are $n$-simplices which are mapped by $g$ linearly onto $\Delta$. Clearly $g_i : \Delta_i^* \stackrel{g}{\to} \Delta$ is an invertible linear map.
Now let $p_i \in \Delta_i^*$ be the unique point such that $g(p_i) = q$. The $p_i$ are contained in the interior of $\Delta_i^*$ (if a $p_i$ would lie on the boundary of $\Delta_i^*$, then also $q = g(p_i)$ would lie on the boundary of $\Delta$).