I think this is done inductively on the skeletons but I can't work out the details.
How do you prove a CW complex is locally path connected
4
$\begingroup$
general-topology
-
0I proved this very carefully here https://math.stackexchange.com/questions/1448789/understanding-construction-of-open-nbds-in-cw-complexes/1462419#1462419 – 2017-06-18
1 Answers
3
You can use the following two general topology facts:
- A disjoint union of locally path-connected spaces is locally path-connected.
- A quotient of a locally path-connected space is locally path-connected (See Lemma 2 in this post).
Let $X_n$ be the n-skeleton of a CW-complex $X$. $X_0$ is obviously locally path-connected. Inductively, $X_n$ is the quotient of the disjoint union of $X_{n-1}$ and n-cells so $X_n$ is locally path-connected. Now $X=\bigcup_{n}X_n$ has the weak topology with respect to the subspaces $X_n$. Equivalently, $X$ has the quotient topology with respect to the map $\coprod_{n}X_n\to X$ defined in terms of the inclusions. Since $\coprod_{n}X_n$ is locally path-connected, so is $X$.