Let $C$ be a triangulation of sphere $S^k$, i.e., $C$ is a geometric simplicial complex and the union of all simplices in $C$ is homeomorphic to $S^k$.
For a vertex $v$ of $C$, $$link_C(v):=\{\sigma\in C: \sigma\cap \{v\}=\emptyset, \sigma\cup\{v\}\in C\}.$$
One claims that $link_C(v)$ is a triangulation of $S^{k-1}$.
I try some examples in the small $k$ cases and feel it is true. But I don't know how to prove the claim. Perhaps there is some topological way to view this problem?
Best Answer
The link is a homology sphere, ie has the homology of a sphere. To see this, denote the link at $v$ by $L$ and the star at $v$ by $S$. Note that $L$ is homotopy equivalent to $S - v$ and both $S$ and $C - v$ are contractible. So $$\tilde H_*(L) \cong \tilde H_*(S - v) \cong H_{* + 1}(S, S - v) \cong H_{* + 1}(C, C - v) \cong \tilde H_{* + 1}(C)$$ where the third isomorphism is by excision. So $L$ is a $(k - 1)$-homology sphere.
To see that the link need not always be a sphere, let $X$ be a homology sphere that is not a sphere (for instance the Poincaré homology sphere). Then its double suspension $\Sigma^2X$ is a sphere, by the double suspension theorem. Now, consider two cases.
$\Sigma X$ is also a sphere. Then, take $C = \Sigma X$, and $v$ to be either suspension point in $\Sigma X$. Then $\text{link}(v) = X$, which by assumption is not a sphere.
$\Sigma X$ is not a sphere. Then, take $C = \Sigma^2 X$, and $v$ to again be either suspension point (now in $\Sigma^2 X$). Then $\text{link}(v) = \Sigma X$, which by assumption is not a sphere.
In either case, we have $C$ is a sphere but $\text{link}(v)$ is not.