Let K be a simplicial complex and σ a simplex of K. Show that the link lk(σ) is a subcomplex of K, if lk(σ) is not empty.

Question

I wanna know if I choose the right way to do my proof? it's correct?

To show that the link $\textrm{Lk}(\sigma)$ is a subcomplex of $\mathcal{K}$, if $\textrm{Lk}(\sigma)$ is not empty, we need to show two things:

1. Every simplex of $\textrm{Lk}(\sigma)$ is a simplex of $\mathcal{K}$.
2. $\textrm{Lk}(\sigma)$ is closed under taking faces, i.e., if $\tau$ is a simplex of $\textrm{Lk}(\sigma)$, then all faces of τ are also simplices of $\textrm{Lk}(\sigma)$.

We begin by recalling the definitions of the link and subcomplex:

• The link $\textrm{Lk}(\sigma)$ of a simplex $\sigma$ in a simplicial complex $\mathcal{K}$ is the set of all simplices $\tau$ in $\mathcal{K}$ such that $\sigma$ and $\tau$ are disjoint, but the union of $\sigma$ and $\tau$ is a simplex of $\mathcal{K}$.
• A subcomplex of $\mathcal{K}$ is a simplicial complex $\mathcal{L}$ that is a subset of $\mathcal{K}$ such that every simplex of $\mathcal{L}$ is also a simplex of $\mathcal{K}$, and every face of a simplex in $\mathcal{L}$ is also a simplex in $\mathcal{L}$.

Now, let's show that the link $\textrm{Lk}(\sigma)$ satisfies these two conditions:

1. Every simplex of $\textrm{Lk}(\sigma)$ is a simplex of $\mathcal{K}$:

Suppose $\tau$ is a simplex in $\textrm{Lk}(\sigma)$. By definition, $\sigma$ and $\tau$ are disjoint, but the union of $\sigma$ and $\tau$ is a simplex of $\mathcal{K}$. This means that $\tau$ is also a simplex of $\mathcal{K}$, since it is a face of a simplex of $\mathcal{K}$.

2. $\textrm{Lk}(\sigma)$ is closed under taking faces:

Suppose $\tau$ is a simplex of $\textrm{Lk}(\sigma)$. We need to show that every face of $\tau$ is also a simplex of $\textrm{Lk}(\sigma)$.

Let $\rho$ be a face of $\tau$. Then $\rho$ is also a face of $\sigma \cup \tau$, since $\sigma$ and $\tau$ are disjoint. Moreover, $\sigma$ and $\tau$ are the only simplices in $\sigma \cup \tau$ that contain $\rho$, since they are the only simplices that contain $\tau$. This means that $\rho$ is a face of $\sigma$ in $\mathcal{K}$, and $\sigma$ is not in $\textrm{Lk}(\rho)$ by definition. Therefore, $\rho$ is a simplex in $\textrm{Lk}(\sigma)$, since it is disjoint from $\sigma$ and their union is a simplex of $\mathcal{K}$. Since $\textrm{Lk}(\sigma)$ satisfies both conditions, it follows that $\textrm{Lk}(\sigma)$ is a subcomplex of $\mathcal{K}$, if $\textrm{Lk}(\sigma)$ is not empty. Thus, the link $\textrm{Lk}(\sigma)$ is a subcomplex of $\mathcal{K}$, if $\textrm{Lk}(\sigma)$ is not empty.

Answer 1

Something seems off in your reasoning in the last paragraph. I would write it as follows:

2. $\textrm{Lk}(\sigma)$ is downward closed:

Let $\tau \in \textrm{Lk}(\sigma)$ be a simplex, with $\rho \subseteq \tau$. We must show that $\rho \in \textrm{Lk}(\sigma)$. We know that $\tau \cup \sigma$ is a simplex in $\mathcal{K}$, and since $\rho \subseteq \tau$, then $\rho \cup \sigma \subseteq \tau \cup \sigma$. So $\rho \cup \sigma$ is a simplex in $\mathcal{K}$. Furthermore, since $\tau$ and $\sigma$ are disjoint, then $\rho$ and $\sigma$ must be disjoint as well, since $\rho \cap \sigma \subseteq \tau \cap \sigma = \varnothing$.

We have shown that $\rho$ satisfies the definition of being a member of $\textrm{Lk}(\sigma)$, therefore $\textrm{Lk}(\sigma)$ is downward closed. Together with 1., this makes $\textrm{Lk}(\sigma)$ a subcomplex of $\mathcal{K}$.