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This is Ex 1.1.19 from Hatcher's algebraic topology.

Show that if $X$ is a path-connected 1-dimensional CW complex with basepoint $x_0$ a 0-cell, then every loop in $X$ is homotopic to a loop consisting of a finite sequence of edges traversed monotonically.

But consider the Hawaiian earring $X$. We can construct a loop $f:[0,1]\to X$, such that $f|_{[1/(n+1),1/n]}$ wraps around $n$th (largest) circle, and $f(0)=x_0$. $f$ is continuous. Isn't this a counter-exmaple to the problem?

JSCB
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    Is the Hawaiian earring a CW complex? I don't think it has the weak (W) topology. – Angina Seng Jul 03 '17 at 14:15
  • From what I have learnt, X having the weaking topology means that for each subset A of X, A is open iff $A\cap X^n$ is open for each n (where $X^n$ is the n skeleton). Can you please tell me explicitly which A fails to satisfy this property? – JSCB Jul 03 '17 at 15:29
  • The weak topology here, in the infinite bouquet $B$ of circles $C_1,\ldots$ attached at a basepoint is that $A$ is open iff $A\cap C_i$ is open in each $C_i$. This bouquet maps continuously and bijectively to the earring $X$ but this is not a homeomorphism. If we delete a point (not the basepoint) from each $C_i$ in $B$ we get an open set; this is not true in the earring, where each neighbourhood of the basepoint contains all bu finitely many of the circles. – Angina Seng Jul 03 '17 at 16:22

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As the Lord Shark pointed out, the Hawaiian Earring is not a CW complex. If it had the weak topology it would be homotopic to the infinite wedge of circles, but the Hawaiian earring has the subspace topology induced by the Euclidean topology. This is exactly what makes the Hawaiian Earring special (not locally simply connected).

EDIT(on the weak topology): It may add to the confusion but to talk about weak topology we must talk about a space $X$ and maps $f:X \rightarrow Y$. Any subspace topology is the coarsest topology such that the inclusion map is continuous and is thus a weak topology with regards to that map. Thus it may be better to distinguish the Hawaiian Earring from a CW complex by noting it is not locally contractable, which a CW complex must be.

EDIT 2 ELECTRIC BOOGALOO To directly address your definition of weak topology: $A$ is open in $X$ iff $A \cap X^n$ is open. This should be tweaked, classically it is stated $ A$ is open iff $A \cap e_i^n$ is open for all $n$-cells $e_i^n$. As Dr. Mosher pointed out this is the strongest (read finest) topology such that the attaching maps are continuous.

Now let $A$ be such that $A \cap e_i^1$ is something like $[0,\frac{1}{2}) \cup (\frac{1}{2},1]$. Then $A$ is open but as $A$ contains the point $(0,0)$ it must contain infinite circles ( there must be an $j$ with $A\cap e_i^1 = e_i^1$ for all $i >j$ which is not the case so we arrive at a contradiction, the space does not have the weak topology ( in CW context.)

Birch Bryant
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  • From what I have learnt, X having the weaking topology means that for each subset A of X, A is open iff $A\cap X^n$ is open for each n (where $X^n$ is the n skeleton). Can you tell me explicitly which A fails to satisfy this property? – JSCB Jul 03 '17 at 15:01
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    Did you mean "not locally simply-connected" instead of "not locally path connected"? Because the Hawaiian Earring is locally path connected. – Stefan Hamcke Jul 03 '17 at 17:24
  • @StefanHamcke yes thank you for catching – Birch Bryant Jul 03 '17 at 17:27
  • Regarding your edit, you are giving the modern standard definition of "weak topology", which unfortunately clashes with the usage of that term in the definition of a CW complex. In this modern terminology, we would instead consider the set of all characteristic maps ${\chi_i : D^{n_i} \to X}_{i \in I}$ (thus mapping to $X$ rather than from $X$), and we would say that the CW topology on $X$ is the strongest topology with respect to which the characteristic maps $\chi_i$ are continuous. – Lee Mosher Jul 03 '17 at 18:41
  • @LeeMosher yes the usage of such terms really gets messy, I had a heck of time figuring the similarities. I suppose a good solution would to have been around when the terms were used in the older sense. But alas I am but a lowly grad student – Birch Bryant Jul 03 '17 at 18:56
  • @LeeMosher it was because of this I shied away from the topology and commented on local properties – Birch Bryant Jul 03 '17 at 18:57
  • Indeed yes, terminology is messy. But the properties are nonetheless clear. – Lee Mosher Jul 03 '17 at 20:09
  • @LeeMosher I made explicit this strongest topology of attaching maps and used this definition to show the space does not have the topology of a CW complex – Birch Bryant Jul 03 '17 at 23:28