- Theorem: \(\pi_1(\bigvee_{i=1}^k S^1) \cong {\Large{*}}_{i=1}^n\ZZ\), the free product of \(n\) copies of \(\ZZ\).
We proceed by induction on \(n\), using the Seifert–van Kampen theorem.
Base case: Take \(n=1\), then \(\pi_1(S^1) = \ZZ\) as proved in Hatcher.
Inductive step: Suppose this holds for all \(k < n\), then we have \(X = \bigvee^n S^1 = \left( \bigvee^{n-1}S^1\right) \vee S^1\). Let \(p\) be the point of common intersection, then let \(U = \bigvee^{n-1} S^1\) and \(V = S^1 \union \theset{p}\).
Then \(U\union V = X\), \(U \intersect V = \theset{p}\), and both \(U,V\) are path-connected. Since we have \(\pi_1(\pt) = 0\), the amalgamated free product reduces to the usual free product. By the induction hypothesis, we have \(\pi_1(U) = {\Large{*}}^{n-1}\ZZ\), so
\[\pi_1(X) = \pi_1(U\cup V) = \pi_1(U) * \pi_1(V) =_{\text{IH}} ({\Large{*}}^{n-1} \ZZ ) * \pi_1(V) = ({\Large{*}}^{n-1} \ZZ) * \ZZ = {\Large{*}}^{n} \ZZ.\]
Definition: Let \(F_n \definedas {\Large{*}^n} \ZZ\) be the free group on \(n\) generators. Lemma: If \(n\neq m, F_n \not\cong F_m\).
If \(F_n \cong F_m\), then abelianizing gives \(\ZZ^n \cong \ZZ^m\). But then tensor both sides with \(\ZZ_2\) over \(\ZZ\), yielding \(\ZZ^n \otimes_\ZZ \ZZ_2 \cong \ZZ^m \otimes_\ZZ \ZZ_2\). But the LHS is isomorphic to \((\ZZ/2\ZZ)^n\), while the RHS is isomorphic to \((\ZZ/2\ZZ)^m\). These are both finite groups: there are 2 elements in \(\ZZ/2\ZZ\), so the first has \(2^n\) elements and the latter has \(2^m\) elements. But if \(2^n=2^m\), then \(n=m\). The lemma follows from the contrapositive.
Now we have all we need - let \(X = S^2 - \theset{p_1, p_2}\) and \(Y = S^3 - \theset{q_1, q_2}\). Then by the previous problems, \(X \homotopic S^1\) and \(Y \homotopic S^2\), so if \(S^2 \cong S^3\) then \(X \homotopic Y\) and \(S^1 \homotopic S^2\). But \(\pi_1(S^1) = \ZZ\) and \(\pi_1(S^2) = 0\), so \(S^1 \not\simeq S^2\), a contradiction.
Solution
1. For \(n\ge1\),\[\boxed{\pi_1\left(\bigvee_{i=1}^nS^1\right)\cong F_n.}\]
Apply van Kampen inductively: adjoining one circle at the common basepoint takes the free product with \(\pi_1(S^1)=\mathbb Z\).
2. If \(F_n\cong F_m\), then \(n=m\).
Abelianization gives \(\mathbb Z^n\cong\mathbb Z^m\), forcing equal ranks.
3. In particular \(S^2\) and \(S^3\) are not homeomorphic.
Removing two points from a sphere preserves homeomorphism type under any hypothetical homeomorphism. But \[S^2\setminus\{p,q\}\simeq S^1,\qquad S^3\setminus\{p',q'\}\simeq S^2.\] Their fundamental groups are \(\mathbb Z\) and \(0\), so the punctured spaces are not homotopy equivalent, hence cannot be homeomorphic.