I'm really not understanding the idea of the (upper) Koszul simplicial complex: For a monomial ideal $I$ and a degree $\mathbf{b}\in\mathbb{N}^n$, define the upper simplicial Koszul complex as $$K^{\mathbf{b}}(I)=\{\text{square free vectors }\tau \mid \mathbf{x}^{\mathbf{b}-\tau}\in I\}.$$ I need to show that this is a cone when $\mathbf{x}^{\mathbf{b}}$ is not the lcm of some of the minimal monomial generators of $I$.
I tried to go about induction on the number of generators, but I wasn't sure what to do in the inductive step.
[For convenience of notations, I will drop all the bolds.]
Presumably you are meant to also assume that $x^b\in I$ (otherwise $K^b(I)$ is empty). Now let $x^a$ the the lcm of all the minimal monomial generators of $I$ which divide $x^b$. By hypothesis, $a<b$; let $i$ be such that $a_i<b_i$. Now observe that if $\tau\in K^b(I)$ and $\tau_i=0$, then $\tau'\in K^b(I)$ as well, where $\tau'$ is $\tau$ modified to have $\tau_i=1$. Indeed, if $x^c$ is any minimal monomial generator of $I$ which divides $x^{b-\tau}$, then $c_i\leq a_i<b_i=b_i-\tau_i$ and so $c\leq b-\tau'$ and $x^c$ still divides $x^{b-\tau'}$. In other words, $i$ can be added to any simplex of $K^b(I)$ to get another simplex; this says exactly that $K^b(I)$ is a cone with vertex $i$.