# Join of simplices and spheres

Suppose $c$ a simplex such that it is the join of two of its faces, say $c=a\ast b$. Let $S$ be defined as $$S=\left\{\sigma\in \partial(c')\big\vert \ \sigma \textrm{ has no vertex in } a\cup b\right\}$$ where $c'$ is the baricentric subdivision of $c$. I would like to show that $S$ is homotopic to a sphere.

Now, this is my idea, that I am not able to complete: a simplex $\sigma$ in $S$ can be written as a sequence of proper inclusions $$\sigma\equiv \alpha_0\subsetneq\ldots\subsetneq\alpha_k$$ such that the $\alpha_i$ are proper faces of $c$ and none of them is contained in $a\cup b$. If I'm not wrong, this space should be equal to the following: take the boundary of $c$ (which is a sphere) and subdivide it, obtaining a sphere again. Now remove $a$ and $b$ (which should be disjoint, as $c$ is the join of them) and "collapse" what you get over the maximum subcomplex that remains. This should be homotopic to a sphere with 2 points removed, so it should be again homotopic to a sphere.

Probably I'm completely wrong, but the other paths I've tried to follow led me to non-sense conclusions.

Consider the set $$T \;=\; \{\sigma \in c' \mid \sigma\text{ has no vertices in }a\cup b\}$$ It is not hard to show that $T$ is homeomorphic with $a \times b$. For example, every line segment with one endpoint in $a$ and one endpoint in $b$ intersects a unique point of $T$, and this defines a map $a\times b \to T$.
Since $a$ and $b$ are simplices, it follows that $T$ is homeomorphic to a ball. The set $S$ you describe is the boundary of this ball, and is therefore homeomorphic to a a sphere.