Let $C$ be a simplicial complex on a finite vertex set and $v$ is a vertex of $C$. The star of $v$ in $C$ is defined as $$st(C,v):=\{\sigma\in C: \sigma\cup\{v\}\in C\}.$$ I feel that $$\tilde{H}_i(st(C,v))=0$$ for all $i$. Is this correct?
Here is my idea to prove it.
Consider the maximal simplices $\sigma_1,..,\sigma_k\in C$ that contain $v$.
The first idea: Let $\tau_i:=\{\sigma\subseteq \sigma_j \text{ for some }1\le j\le i\}$. We have $\tau_k=st(C,v)$ and we prove by induction on $i$ that the reduced homology group of $\tau_i$ vanishes.
When $i=1$, the statement holds. For larger $i$, let $B=\{\sigma\subseteq \sigma_i\}$. Then $\tau_{i-1}\cap B$ is the power set of a subset of $\sigma_i$, which is contractible. Therefore by Mayer–Vietoris sequence, the reduced homology group of $\tau_i=\tau_{i-1}\cup B$ vanishes.
The second idea: (Something essentially same as the first one. can be ignored if the first idea is correct.)
We divide them into different components $C_1,...,C_\ell$: $\sigma_i$ and $\sigma_j$ are in the same component if there is a sequence $\sigma_{i_1},,,,\sigma_{i_t}$ such that $\sigma_i=\sigma_{i_1}, \sigma_{j}=\sigma_{i_t}$, and $|\sigma_{i_a}\cap\sigma_{i_{a+1}}|\ge 2$. (Note that $v$ is in each of the simplices.)
Then the star is the wedge sum of these components at $v$. Since the reduced homology group of the star is the directed sum of the reduced homology groups of these components, it is enough to prove that the reduced homology group of each component is 0.
Let $C_p$ be a component. We might be able to add the simplices one by one (I am not sure about this step) so that by induction and Mayer–Vietoris sequence, if $\sigma,\tau,\sigma\cap \tau$ have zero reduced homology group, so is $\sigma\cup \tau$. (If we can add one by one, note that by maximality, each subset of the newly added simplex is in the star, that is why it has zero homology.)