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