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I've found few different formulations for Van-Kampen's theorem the first of which states that if $X = A \cup B$ where $A$ and $B$ are path-connected, contain the basepoint $x_0$ and $A \cap B$ is path-connected then $$\pi_1(X) \cong \pi_1(A) \ast \pi_1(B)/N$$ where $N$ is the smallest normal subgroup of $\pi_1(A) \ast \pi_1(B)$ with the identification $\iota_1([\gamma])=\iota_2([\gamma])$ for $\gamma \in A \cap B$ and $\iota_1: \pi_1(A \cap B) \to \pi_1(A), \iota_2:\pi(A \cap B) \to \pi_1(B)$.

The second one is done with amalgamation which is essentially the same thing, but they just denote $\pi_1(X) \cong \pi_1(A) \ast \pi_1(B)/N$ as $\pi_1(X) \cong \pi_1(A) \ast_{\pi(A \cap B)} \pi_1(B)$.

My question is why do we need this normal subgroup $N$ here at all? And why do we require it to be the smallest one? I see that we want to identify the loops in $\pi_1(A \cap B)$ which are sent by the inclusions $\iota_1$ and $\iota_2$ to $\pi_1(A)$ and $\pi_1(B)$, but why do we want to do this kind of thing?

Emil Grönberg
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1 Answers1

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Let me steal this diagram from Wikipedia:

enter image description here

It's clear that:

  • $\pi_{1}(U_1 \cap U_2)$ maps to $\pi_{1}(U_1)$ and $\pi_{1}(U_2)$. This is the map on homotopy induced by inclusion.
  • $\pi_{1}(U_1)$ and $\pi_{1}(U_2)$ map to $\pi_{1}(X)$. This is also given by inclusion.

It's not hard to see:

  • Every element of $\pi_{1}(X)$ can be factored into a product of elements in $\pi_{1}(U_1)$ and $\pi_{1}(U_2)$. This is a standard homework exercise in a first topology course.
  • As a result, there is a surjection from $\pi_{1}(U_1) \ast \pi_{1}(U_2)$ to $\pi_{1}(X)$.
  • However, this surjection will generally not be injective. Suppose that there is a nontrivial element $\gamma \in \pi_{1}(U_1 \cap U_2)$ that remains nontrivial when mapped to $\pi_{1}(X)$. Then the two, distinct nontrivial elements $i_{1}(\gamma) \ast 1$ and $1 \ast i_{2}(\gamma)$ in $\pi_{1}(U_1) \ast \pi_{1}(U_2)$ map to the same element of $\pi_{1}(X)$.
  • To fix this, we can identify all pairs of the form $i_{1}(\gamma) \ast 1$ and $1 \ast i_{2}(\gamma)$.
  • In the language of group theory, if you want to make such identifications, you introduce the relators $(i_{1}(\gamma) \ast 1)^{-1}(1 \ast i_{2}(\gamma))$.
  • Adding group relations is the same thing as quotienting out by smallest normal subgroup (the normal closure) containing the relators.

The remaining work to be done:

  • We can see that there is a surjection from the amalgamated product to $\pi_{1}(X)$. But it is injective (and hence a homeomorphism)? In other words, has our amalgamation trick taken care of all possible failures of injectivity? Working this out completes the proof, and is a more difficult.
Elchanan Solomon
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  • In the first bullet on "It's not hard to see" what do you mean by factoring into a product? Since $X = U_1 \cup U_2$ do we have some way to use the product of $U_1$ and $U_2$? – Emil Grönberg Jul 19 '22 at 08:43
  • Not the product of the the sets, the product of the fundamental groups. Think of the simple case when $U_1 \cap U_2$ is just a point ${p}$. Then any loop in $X$ starting at $p$ moves around in $U_1$, then passes through $p$ and winds around in $U_2$ for a bit, then hits $p$ again and moves into $U_1$, etc. Thus the whole loop is a product of loops, each of which is entirely in $U_1$ or $U_2$. – Elchanan Solomon Jul 19 '22 at 08:47