# Seifert-van Kampen theorem, classical version

This is from Munkres' Topology page 431:

Theorem 70.2 (Seifert-van Kampen theorem, classical version). Assume the hypotheses of the preceding theorem. Let

$$j : \pi_1(U, x_0) * \pi_1(V, x_0) \longrightarrow \pi_1(X, x_0)$$

be the homomorphism of the free product that extends the homomorphisms $$j_1$$ and $$j_2$$ induced by inclusion. Then $$j$$ is surjective, and its kernel is the least normal subgroup $$N$$ of the free product that contains all elements represented by words of the form

$$(i_{1}(g))^{-1} i_{2}(g)),$$

for $$g \in \pi_{1}(U \cap V, x_{0}).$$

and by hypotheses of the preceding theorem it means that $$X$$ is a topological space and $$U$$, $$V$$ open subsets of $$X$$ such that $$X = U \cup V$$ and $$x_0 \in U \cap V$$

$$j_1:\pi_1(U,x_0)\to\pi_1(X,x_0)$$ and $$j_2:\pi_1(V,x_0)\to\pi_1(X,x_0)$$ are inclusion induced homomorphisms.

I read the proof and understood it but something is really bothering me and that is how do we know that the homomorphism $$j$$ mentioned in the theorem even exists?

• This comes down to understanding the definition of the free product. Mar 25 at 15:42
• can you elaborate on it a bit? Mar 25 at 15:45
• I'd start here: en.wikipedia.org/wiki/Free_product Mar 25 at 15:46
• I've read the definition before so if you could explain how this comes down to understanding the definition it would be appreciated Mar 25 at 15:49

It tells you: $$j_1:U\subset X$$ and $$j_2:V\subset X$$ induce maps on $$\pi_1$$, $$\pi_1(U)\to\pi_1(X)$$ and $$\pi_1(V)\to\pi_1(X)$$, and these are genuine homomorphisms, and if you have two groups $$A,B$$ and a third group $$C$$ and homomorphisms $$A\to C,B\to C$$ then there is a unique associated homomorphism $$A\ast B\to C$$. In this case with $$A=\pi_1(U),B=\pi_1(V),C=\pi_1(X)$$, there is a unique homomorphism $$j:\pi_1(U)\ast\pi_1(V)\to\pi_1(X)$$ associated to $$j_1,j_2$$.