Answer: Fundamental Group of S¹ ∨ S¹ Is Non-Abelian

Key Idea / Intuition

The wedge sum $S^1 \vee S^1$ looks like the letter "∞" — two loops sharing a basepoint. A loop in this space is a word made of "go around the first circle" and "go around the second circle" in any order. There is no reason these two generators should commute: going around $a$ then $b$ traces a genuinely different path than $b$ then $a$, and there is no homotopy in the space that could swap them. Van Kampen's theorem makes this precise: when two open sets meet in a simply-connected overlap, the fundamental group is the free product (not direct product) of their fundamental groups.

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Formal Proof / Solution

Setup. Let $p \in S^1 \vee S^1$ be the wedge (base) point. Write the two circles as $A$ and $B$ with $A \cap B = \{p\}$.

Applying the Seifert–Van Kampen Theorem.

Choose open sets: $U = A \cup (\text{small open arc of } B \text{ near } p), \quad V = B \cup (\text{small open arc of } A \text{ near } p).$

Then:

• $U \simeq S^1$, so $\pi_1(U) \cong \mathbb{Z}$, generated by $a$ (loop around $A$).
• $V \simeq S^1$, so $\pi_1(V) \cong \mathbb{Z}$, generated by $b$ (loop around $B$).
• $U \cap V \simeq$ a small cross-shaped neighborhood of $p$, which is contractible, so $\pi_1(U \cap V) = \{e\}$.

Van Kampen's theorem states: $\pi_1(X) \cong \pi_1(U) *_{\pi_1(U \cap V)} \pi_1(V) = \mathbb{Z} *_{\{e\}} \mathbb{Z} = \mathbb{Z} * \mathbb{Z}.$

Since the amalgamation is trivial (over the trivial group), the result is the free product, i.e., the free group on two generators.

The result: $\pi_1(S^1 \vee S^1) \cong \mathbb{Z} * \mathbb{Z} = F_2,$ the free group on generators $a$ and $b$.

Explicit non-commuting loops.

Let $\alpha$ be the loop that goes once around circle $A$, and $\beta$ the loop that goes once around circle $B$ (both based at $p$).

• The loop $\alpha \cdot \beta$: first traverse $A$, then traverse $B$.
• The loop $\beta \cdot \alpha$: first traverse $B$, then traverse $A$.

These are not homotopic: in $F_2$, the word $ab \neq ba$. Topologically, no continuous deformation can swap the order, because any homotopy would need to "slide" one loop past the other, but they are only connected at the single point $p$ — there is no room to slide.

Why does abelianness fail despite each piece being abelian?

The free product $G * H$ is abelian only if one of $G, H$ is trivial. Even though $\mathbb{Z}$ is abelian, the free product $\mathbb{Z} * \mathbb{Z}$ has elements like $aba^{-1}b^{-1} \neq e$. The abelianization of $F_2$ is $\mathbb{Z} \times \mathbb{Z}$ (the direct product), which corresponds to $H_1(S^1 \vee S^1) \cong \mathbb{Z}^2$ — but homology loses the non-commutativity that $\pi_1$ captures.

Moral: The free product glues generators with no relations between the two groups. The direct product would impose $ab = ba$ for all $a \in G, b \in H$ — but Van Kampen only introduces relations coming from $\pi_1(U \cap V)$, and when that is trivial, no such relations are forced.