# Apparent Implication That $\pi_1(S^1, x_0)$ is Trivial

I know this isn't true, but I can't seem to figure out why. In my topology class, we recently went over the theorem in Munkres section 59 stating that if $$X = U \cup V$$, with $$U, V$$ open in $$X$$, and $$U \cap V$$ path-connected with $$x_0 \in U \cap V$$, then if $$\iota \colon U \hookrightarrow X$$ and $$\jmath \colon V \hookrightarrow X$$ are the respective inclusion maps, then the fundamental group $$\pi_1(X, x_0)$$ is generated by the images of $$\pi_1(U, x_0)$$ and $$\pi_1(V, x_0)$$ under the induced homomorphisms $$\iota_*, \,\, \jmath_*$$.

This obviously implies that when $$\iota, \jmath$$ are the identity map, that $$\pi_1(X, x_0)$$ is trivial. But this is where I run into a problem. $$S^1$$ is an open subset of itself, so let $$X = U = V = S^1$$. Then since $$U \cap V = S^1$$ is path-connected, and the inclusion maps are both just the identity map on $$S^1$$, this implies that $$\pi_1(S^1, x_0)$$ is trivial, a clear contradiction. But I can't find the hole in my reasoning. I'm sure it's something stupid and simple, but could someone point out what I did wrong?

• The inclusion maps are both the identity and what that implies is that $\pi_1(S^1)$ is generated by $\pi_1(S^1)$, which is vacuously true. The identity isn't the zero map! Commented Dec 8, 2020 at 6:45

## 2 Answers

The induced homomorphism of fundamental groups is not trivial for the identity map of $$S^1$$. It corresponds to the element $$1$$ in $$\pi_1(S^1)\cong\Bbb Z$$.

• Wow, I feel very foolish for missing that. Thanks! Commented Dec 8, 2020 at 6:49
• Don't feel bad. You're supposed to play around with it a bit.
– user403337
Commented Dec 8, 2020 at 6:53
• @fluentsandfluxions ditto what Chris said. I would encourage you to try computing $\pi_1(S^1)$ using van Kampen in the other "obvious" way to see what goes wrong. I don't actually think this computation can be done via van Kampen. Commented Dec 8, 2020 at 16:57
• I don't know if you can do Mayer-Vietoris, it's been a while... But that's another thought.
– user403337
Commented Dec 8, 2020 at 20:57
• Looking it up, I see that Mayer-Vietoris is more for homology.
– user403337
Commented Dec 8, 2020 at 21:07

Just to be a bit more explicit than the other answers.

The exact statement is that $$\pi_1 (X) = \pi_1(U) * \pi_1(V) /N$$ where $$N$$ basically describes how the loops of $$U \cap V$$ fits inside $$U$$ and inside $$V$$, then equates both of those.

So in your particular example we have that $$\pi_1(U) = < a>$$ as it's just the circle. $$\pi_1(V) = < b>$$ as well, and finally $$\pi_1(U \cap V) = $$.

The inclusion maps are identity as you remarked so $$c$$ gets mapped to $$a$$ and $$c$$ gets mapped to $$b$$ respectively. Since these are the generators this fully describes everything.

So we finally get that $$\pi_1(X) = */ = = (or ) = \mathbb{Z}$$

• Thank you very much for the in-depth explanation! Commented Dec 8, 2020 at 6:50