# In what sense are trivalent graphs “generic” 1-dimensional CW-complexes?

The Wikipedia article CW complex mentions that "trivalent graphs can be considered as generic 1-dimensional CW complexes", offering this explanation:

Specifically, if $X$ is a 1-dimensional CW complex, the attaching map for a 1-cell is a map from a two-point space to $X$, $f : \{0,1\} \to X$. This map can be perturbed to be disjoint from the 0-skeleton of $X$ if and only if $f(0)$ and $f(1)$ are not 0-valence vertices of $X$.

There is a similar statement in the article Cubic graph: "if one considers a graph to be a 1-dimensional CW complex, cubic graphs are generic in that most 1-cell attaching maps are disjoint from the 0-skeleton of the graph".

Could someone help me to follow these explanations? I'm still not sure what is meant by the term "generic". Does it express some kind of universal property of trivalent graphs among 1-dimensional CW complexes?

I think you may simply consider trivalent graphs as $1$-dimensional CW complexes such that exactly three edges ($1$-cells) are emanating from each vertex $(0$-cells).

I will try to explain the meaning of "generic map" in the following.

Consider the following simple example: If two straight lines are placed in a plane, then there are three possible situations:

(i) They coincide with each other.

(ii) They intersect transversely at one point.

(iii) They are parallel and disjoint.

We say (ii) is the "generic" situation, and (i) and (iii) are not generic. Any small perturbation can turn the situation (i) and (iii) into (ii), but (ii) remains unchanged under any small perturbation. In other words, "$1$-dimensional objects in a $2$-dimensional space generically intersect transversely at $0$-dimensional subspaces".

Similarly, you can see that $0$-dimensional objects in a $1$-dimensional space are generically disjoint. Thus the image of an attaching map$$f: \{0, 1\} \to X$$ of a $1$-cell to a $1$-dimensional CW complex $X$ (a graph) is generically disjoint from any vertices of $X$. After attaching the $1$-cell the points $f(0)$ and $f(1)$ become new vertices of $X \cup (\text{the }1\text{-cell})$, that are trivalent vertices.

• That was a very helpful explanation, thanks! – Noam Zeilberger Jul 25 '17 at 8:09