I'm doing exercises about simplicial complexes and I'm stuck with one for which I'll first give some definitions.
Let $K$ a simplicial complex and $v\in K^0$ a $0$-simplex (vertex).
The star of $v$ in $K$, $st(v;K)=\{\sigma\in K\mid \exists \tau\leq\sigma: v\in\tau\}$
The link of $v$ in $K$, $lk(v;K)=\{\sigma\in st(v;K)\mid v\notin\sigma\}$.
The deletion of $v$ y $K$, $del(v;K)=\{\sigma\in K\mid v\notin\sigma\}$.
The problem is the following:
Show that if there is some $v\in K^0$ such that $lk(v;K)$ is collapsible, then $K$ collapses to $del(v;K)$.
The definition of collapse is a bit long, so I'll leave it here if necessary.
I've been thinking of collapsing the star to the link using that the star is the simplicial cone over the link with respect to $v$. For each maximal simplex $\sigma$ and free face $\tau<\sigma$ of $lk(v;K)$ I can get a maximal simplex $v\sigma$ and a free face $v\tau<v\sigma$ of $st(v;K)$, hence I can make a collapse here.
The problem is that I don't see how to complete the proof. Once I perform this collapse, the new link is the first collapse of the original link, so I guess I should do parallel collapses in $lk(v;K)$ and in $K$ redefining the link of $v$ in $K$ at each step (the original link will remaing untouched in $K$).
I feel if I keep doing this, I'll end with $v$ conected by a $1$-simplex to a vertex of the original $lk(v;K)$, and then I can make the final collapse. An example of this would be the following picture where $K$ is a $3$-simplex. 