Understanding the $\operatorname{link}$ of a simplex $\sigma$ I have a question that goes as follows:

Let $K$ be an abstract simplicial complex and $\sigma \in K$. Define the link of $\sigma$ as follows:
$$\operatorname{link}_\sigma(K) := \{\tau \in K | \tau \cap \sigma = \emptyset \, ,\, \tau \cup \sigma \in K\}$$
Show that the link is a simplicial complex.

Then, as a follow-up question,

Let K be the octahedron. Draw the link of $\sigma$ when $\sigma$ is
(i) a vertex
(ii) an edge
(iii) a 2-simplex.

I don't know how to get started. Can anyone advise on what the proof conditions for a simplicial complex are, and how to go about proving them?
Also, if anyone has a more intuitive explanation on what a link is to help me to picture and draw the link of $\sigma$, I would be very grateful!
 A: There is a nice picture of what links are on Wikipedia:

The link of a simplex $\sigma$ is the set of simplices $\tau$ that don't intersect $\sigma$ but such that $\sigma \cup \tau$ is also a simplex. This is of course identical to the definition translated in English, but hopefully it helps.
So for the octahedron. Your first task should be to draw one:

and then number the vertices from $1$ to $6$. Let's say $1$ is the vertex on the left, $2,3,4,5$ are the vertex of the middle square in the clockwise order, and $6$ is the vertex on the right. Then you can write down exactly what the simplices are in $K$. It's a bit of a daunting task: there are 6 vertices, 12 edges, and 8 faces, so in total $1+6+12+8=27$ simplices (don't forget the empty set).
Now you want to find the links.
Let's start with the link of a vertex. Look at the vertex $1$ on the left for example. You're looking for simplices $\tau$ that 1. don't contain the vertex on the left ($1 \not\in \tau$), and 2. when you take the union $\{1\} \cup \tau$, this is still a vertex of $K$.


*

*For example, does the vertex on the right belong to the link? In other words, is $\{1\} \cap \{6\} = \varnothing$ and $\{1, 6\} \in K$? The first condition is OK, but the second condition isn't because there is no edge between $1$ and $6$. For the four other vertices, the two conditions are OK, because there is an edge from $1$ to the vertex. So in $\operatorname{lk} \{1\}$, the vertices are $\{2,3,4,5\}$.

*What about the edges? Does an edge between two elements of $\{2,3,4,5\}$ belong to the link? Yes: if you take such an edge and add the vertex $1$, then this is a simplex of the octahedron: there are faces $\{1,2,3\}, ${1,3,4} etc. So the whole square in the middle of the octahedron is included in the link of $1$.

*There are no $2$-faces between the vertices of $\{2,3,4,5\}$ so we are done: the link of the vertex on the right is the square in the middle, between the orange/pink and green parts.
Now hopefully you can try to find the links of a edge and of a face. It's the kind of exercise that you don't "get" if you don't do it yourself so I'll leave you to it.
