Given two connected open sets $U,V \subset X$ such that $U \cap V$ is path connected and $U \cup V = X$, then $\pi_1(U) \ast_{\pi_1(U \cap V)} \pi_1(V) \cong \pi_1(X)$. This is of course the Seifert Van Kampen theorem, a question in Munkres asks if the homomorphism induced by the inclusion map of $i: V \rightarrow X$ is trivial what can you say about the homomorphism induced by $j: U \rightarrow X$.

It's clear to me that $j_\ast$ must be surjective and therefore $\pi_1(X) \cong \pi_1(U)/ker(j_\ast)$. My question is can you say anything more? Does the kernel of $j_\ast$ relate at all to $F$, as given in the usual diagram. Would the kernel be at all related to the normal closure of the image of the induced homomorphism of the inclusion $k: U \cap V \rightarrow U$?


Notice that the question is purely algebraic. If $G=\pi_1(U)$, $G'=\pi_1(V)$ and $H=\pi_1(U\cap V)$, you are asking what is the kernel of the obvious map $G\to G*_HG'$ (where the amalgamation happens along the images of $H$ in $G$ and $G'$) if the map $G'\to G*_HG'$ is trivial.

  • $\begingroup$ So it would be the normal closure of $H$ in $U$? $\endgroup$ – JSchlather Mar 3 '11 at 16:50
  • $\begingroup$ The kernel is normal and contains $H$, so it must contain its normal closure. How do you do the other containment? :) $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '11 at 22:04

Indeed, $j_*$ must be surjective. There are certainly many possibilities for the kernel:

(a) $j_*$ could be the zero map, for instance if $X$ is simply connected ($U$ need not be simply connected here);

(b) $j_*$ could be injective, for instance if $V$ is a point;

(c) or it could be somewhere in between, for instance if $X$ is a torus, $U$ a punctured torus, $V$ a disk, and $U\cap V$ a circle. (Here $j_*$ is the "abelianization" map from $F_2$ to $Z^2$.)

There must be many other simple examples. I would guess that Munkres is looking for "$j_*$ is surjective."

  • $\begingroup$ Right, there are many different possibilities for the kernel of $j_\ast$, but I don't see how that implies that you can't describe the kernel of $j_\ast$ in terms of $U \cap V$. $\endgroup$ – JSchlather Mar 3 '11 at 18:49

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