I believe I have a counterexample to Problem 5-17 in John Lee's Introduction to Topological Manifolds. Can someone please confirm that it is a counterexample or explain why it isn't? Here is the statement of the problem.
"Suppose $\sigma = [v_0, \ldots, v_k]$ is a simplex in $\mathbb{R}^n$ and $w \in \mathbb{R}^n$. If $\{w, v_0, \ldots, v_k\}$ is an affinely independent set, we say that w is affinely independent of $\sigma$. In this case, the simplex $[w,v_0, \ldots, v_k]$ is denoted by $w \ast \sigma$ and is called the cone on $\sigma$. More generally, suppose $K$ is a finite Euclidean simplicial complex and $w$ is a point in $\mathbb{R}^n$ that is affinely independent of every simplex in $K$. Define the cone on $K$ to be the following collection of simplices in $\mathbb{R}^n$: $w \ast K = K \cup \{[w]\} \cup \{ w \ast \sigma : \sigma \in K \}$. Show that $w \ast K$ is a Euclidean simplicial complex whose polyhedron is homeomorphic to the cone on $|K|$."
My counterexample is motivated by Munkres' definition of the cone on $K$, in which it is required that the ray from any point $p$ in $K$ to $w$ intersect $K$ only at the point $p$. Let $e_0=0$, and let $(e_1,e_2)$ be the standard basis for $\mathbb{R}^2$. Consider the boundary of the standard $2$-simplex $\Delta_2 = [e_0, e_1, e_2]$, $\partial \Delta_2 = \{[e_0,e_1], [e_0,e_2], [e_1,e_2]\}$, which is a $1$-dimensional Euclidean simplicial complex in $\mathbb{R}^2$. Let $w=(1,1)$. Then $w$ is affinely independent of every simplex in $\partial \Delta_2$, but, for example, $w\ast[e_0,e_1] \cap [e_1,e_2]$ is not a face of either simplex, as it is the lower half of $[e_1,e_2]$.
Any help would be much appreciated.
(Addition -- 5/26/2017) Thank you guys for your comments. It seems that the additional assumption that, for any $\sigma \in K$, $(w \ast \sigma) \cap |K| = \sigma$ is needed, where $|K|$ is the polyhedron of $K$. It is sufficient, and here I think is a brief argument that it is necessary. Suppose $w \ast K$ is a Euclidean simplicial complex. Let $q \in (w \ast \sigma) \cap |K|$, say $q \in \tau$, where $\tau \in K$. Then $q \in (w \ast \sigma) \cap \tau \neq \varnothing$, so $(w \ast \sigma) \cap \tau$ is a face of both $\tau$ and $w \ast \sigma$. The only faces of $\tau$ are simplices in $K$, and the only faces of $w \ast \sigma$ that are simplices in $K$ are faces of $\sigma$, so $q$ is in some face of $\sigma$. Thus $(w \ast \sigma) \cap |K| \subset \sigma$, and the other inclusion is immediate. The counterexample then shows that this additional assumption isn't implied by the existing assumptions.
