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This is the problem 1.2.18 from Hatcher's algebraic topology. Let $X=\{0,1,1/2,1/3,1/4,...\}$, inheriting a subspace topology from the real line, and $SX$ be its suspension, which looks like this:

enter image description here

We need to show that its fundamental group is free with a countable set of generators.

I tried Van Kampen theorem, but cannot find a good cover. Please give me some hints. Thank you.

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  • $\begingroup$ Maybe give a reference for the text, too. Not just the exercise. ;) On a related note, you should be able to find a homotopy equivalent space (Hawaian earings) whose fundamental group you know. $\endgroup$
    – Tyrone
    Jul 10, 2017 at 13:58
  • $\begingroup$ Oops, sorry for that incomplete sentence. By the way, I don't think $SX$ is homotopy equivalent to the Hawaian earring, since the later has an uncountable fundamental group. $\endgroup$
    – JSCB
    Jul 10, 2017 at 14:01
  • $\begingroup$ Collapse the vertical map on the left hand side of your diagram to a point to get a wedge of circles. Just be careful with how the resulting quotient is topologised. $\endgroup$
    – Tyrone
    Jul 10, 2017 at 14:37
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    $\begingroup$ @Tyrone So you're saying to calculate $\Sigma X$. How does that help in calculating $SX$? $\endgroup$
    – freakish
    Jul 10, 2017 at 14:45
  • $\begingroup$ $SX$ is homeomophic to sum of circles of radius $\frac{1}{\pi}\sqrt{(\frac{1}{n})^2+\frac{1}{2})^2}$ for $n=1,2,\cdots$. $\endgroup$
    – Andrews
    Nov 29, 2019 at 19:05

1 Answer 1

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Let $X=\{x_0, x_1, \ldots\}$ be a countable discrete space.

Take $a_0 = [(x_0, 1)]\in SX$ to be a basepoint (the choice of $x_0$ doesn't matter, they all give the same point in $SX$).

Now for each $i\in\mathbb{N}$ define

$$V_i=\bigg\{(x_i, t)\ \bigg|\ \frac{1}{3} < t \leq 1\bigg\}$$

Note that each $V_i$ is open in $X\times[0,1]$. If we now go to the suspension then we can define

$$U_i=p\big(\{x_i\}\times[0,1]\big)\cup \bigcup_i p\big(V_i\big)$$

where $p:X\times[0,1]\to SX$ is the projection. So it's like a circle with a countable umbrella at the top.

A little bit of gymnastics is needed to show that each $U_i$ is open in $SX$ and is homotopically equivalent to $S^1$. Also any interesection of the form $U_{i_1}\cap\cdots\cap U_{i_n}$ is contractible (as long as there are at least two different $U_i$ there). Every such intersection is an umbrella without full circle, you can contract it to the basepoint.

Thus van Kampen applies and therefore

$$\pi_1(X)\simeq *\pi_1(U_i)\simeq *_i\pi_1(S^1)\simeq *_i\mathbb{Z}$$

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  • $\begingroup$ I don't think my $X$ is discrete since it has limit point $0$. Maybe I forgot to mention that $X$ get a subspace topology from the real line. $\endgroup$
    – JSCB
    Jul 10, 2017 at 14:20
  • $\begingroup$ @ᴊᴀsᴏɴ Oh, right, I missed that. This makes things more difficult. Let me think about it. Anyway I'll leave the answer for discrete case. $\endgroup$
    – freakish
    Jul 10, 2017 at 14:22
  • $\begingroup$ @ᴊᴀsᴏɴ Hm, here's one thought. If $X$ is your space and $Y$ is a discrete countable space then take any bijection $f:X\to Y$ which induces a bijection $F:SX\to SY$. None of them are continous however I think that $\pi_1(F)$ is well defined anyway because loops in $SX$ are very specific, they cannot be "too big" in the sense that the umbrella I've defined above cannot be in the image of any loop. I'm not 100% sure but maybe this can be helpful to you. $\endgroup$
    – freakish
    Jul 10, 2017 at 14:54
  • $\begingroup$ See if this is OK: Let (0,1) be the base point of SX. Let U be a nbd of (0,1). For each path c in SX, there must exist a finite subset W of X such that c lies in the union of U and SW. In fact we can deform c so that c just lie in SW. The restriction of F on SW is continuous, so it induces a well defined map. This concludes that $F_*$ is well defined. $\endgroup$
    – JSCB
    Jul 10, 2017 at 15:54

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