I think this is done inductively on the skeletons but I can't work out the details.
$\begingroup$
$\endgroup$
8
-
4$\begingroup$ They're not just locally path-connected, they're locally contractible. There's a key theorem about CW-complexes, that the inclusion of any of any subcomplex into the entire CW-complex is a cofibration. Look at that proof and the neighbourhoods constructed in that proof. That should give you the idea for how to prove what you want to prove. $\endgroup$– Ryan BudneyCommented Jun 19, 2011 at 5:56
-
1$\begingroup$ Please give full details in the question, not just the title. Also, what is a CW complex? $\endgroup$– Asaf Karagila ♦Commented Jun 19, 2011 at 6:00
-
5$\begingroup$ @Asaf What additional details do you want, exactly? And all definitions are easily googlable. $\endgroup$– Grigory MCommented Jun 19, 2011 at 6:07
-
1$\begingroup$ @Grigory: It's not that I complain about lack of definitions. I complain about bad formatting, while at it I was asking what is a CW complex. $\endgroup$– Asaf Karagila ♦Commented Jun 19, 2011 at 6:13
-
1$\begingroup$ @Asaf: First google hit: CW complex: A space obtained by gluing disks together. The topologist's preferred notion of a polyhedron. All reasonable (geometric) spaces are CW complexes. C: closure finite, W: weak topology. Inventor: J.H.C. Whitehead. $\endgroup$– t.b.Commented Jun 19, 2011 at 11:12
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
6
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 $X$.
-
$\begingroup$ You also need finiteness, in terms of number of cells. For example, consider the CW complex constructed starting with the unit square sitting in the first quadrant of the plane. Now add horizontal lines between the points (0,1/2) & (1,1/2), and between (0,1/3) & (1,1/3), and between (0,1/4) & (1,1/4), etc. Now the interior of the base of the square on the x-axis is not locally path connected. $\endgroup$– PossumPCommented Sep 18, 2020 at 14:47
-
2$\begingroup$ That's not a CW-complex. By definition, a CW-complex has the weak topology with respect to it's cells. $\endgroup$ Commented Sep 18, 2020 at 15:26
-
$\begingroup$ You can also do this without induction on skeleta. Choosing $\chi_i : D_i \to X$ to be an indexed set of characteristic maps of the cells of $X$ (of all dimensions), it follows that $X$ has the quotient topology with respect to the disjoint union map $\chi : \sqcup_i D_i \to X$. $\endgroup$ Commented Dec 31, 2023 at 14:35
-
1$\begingroup$ @PetraAxolotl The topologist's sine curve is connected and therefore in particular not the disjoint union of two non-empty spaces. $\endgroup$ Commented Mar 31 at 21:10
-
1$\begingroup$ @BenSteffan Got it! Thanks for the clarification. $\endgroup$ Commented Mar 31 at 21:45