Theorem. Let $X$ be the union of two path-connected open sets $A$ and $B$ and assume that $A\cap B\neq \emptyset$ is simply-connected. Let $x_0$ be a point in $A\cap B$ and all fundamental groups will be written with respect to this base point. Let $\Phi:\pi_1(A)\sqcup \pi_1(B)\to \pi_1(X)$ be the natural homomorphism induced from the maps $\pi_1(A), \pi_1(B)\to \pi_1(X)$ (Here '$\sqcup$' denotes the free product). Then $\Phi$ is and isomorphism.

(The above theorem is given in more general form in Hatcher's Algebraic Topology, but for me the above special case suffices.)

I can see that $\Phi$ is surjective. So we need to address the injectivity of $\Phi$.

Write $G=\pi_1(A)\sqcup \pi_1(B)$. Suppose $\gamma$ is a loop in $A$ based at $x_0$. Think of $[\gamma]$ as member of $G$. Assume that $\Phi([\gamma])$ has the trivial homotopy class in $X$. In order for $\Phi$ to be injective, it is necessary that $[\gamma]$ be the identity element of $G$.

My question is whether or not the following statement is correct:

Statement. $[\gamma]$ is the identity element of $G$ if and only if $\gamma$ has trivial homotopy class in $\pi_1(A)$.

Please check my last statement. For if the above is wrong then it would mean I have to go back to free products.)

  • $\begingroup$ This statement is correct, but it is unclear to me how this is useful to show that $\Phi$ is injective, since most elements of $G$ are not just elements of $\pi_1(A)$. $\endgroup$ – Eric Wofsey May 21 '16 at 7:31
  • $\begingroup$ @EricWofsey You are right. The statement as such does not do much. Through this post I am merely trying to confirm that I am getting things right. I have been confused by free products in the past. Reading van Kampen has made it necessary for me to understand them. $\endgroup$ – caffeinemachine May 21 '16 at 7:38

If I am understanding you correctly, you are asking whether the canonical inclusion map $\pi_1(A)\to \pi_1(A)\sqcup\pi_1(B)$ is injective. This is true, and follows from the concrete description of elements of the free product as reduced words. Explicitly, an element of $\pi_1(A)\sqcup\pi_1(B)$ is a finite sequence $(a_1,a_2,\dots, a_n)$ where the $a_i$ alternate between being non-identity elements of $\pi_1(A)$ and non-identity elements of $\pi_1(B)$. The inclusion map $\pi_1(A)\to\pi_1(A)\sqcup\pi_1(B)$ then sends $a\in\pi_1(A)$ to the sequence of length $1$ consisting only of $a$ if $a$ is not the identity, and sends the identity to the sequence of length $0$ (which is the identity of $\pi_1(A)\sqcup\pi_1(B)$). This map is obviously injective.

  • $\begingroup$ Thank you. This clarifies things. The topological interpretation of the statement is nice though. If one can deform a loop in $A$ to a point by accessing the entirity of $X$, then one can do it even when one is constrained to remain in $A$! Am I right? $\endgroup$ – caffeinemachine May 21 '16 at 7:42
  • $\begingroup$ That's correct. $\endgroup$ – Eric Wofsey May 21 '16 at 7:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.