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

Question

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?

Answer 1

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

Answer 2

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 $\pi_1(U \cap V) = <c>$.

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 $\pi_1(X) = <a> *<b>/<ab^{-1}> = <a, b: ab^{-1}> = <a> (or <b>) = \mathbb{Z}$