# The Fundamental Group of $S^n$

$\textbf{Theorem 59.1.}$
Suppose $X = U \cup V$, where $U$ and $V$ are open sets of $X$. Suppose that $U \cap V$ is path connected, and that $x_0 \in U \cap V$. Let $i$ and $j$ be the inclusion mappings of $U$ and $V$, respectively, into $X$. Then the images of the induced homomorphisms
\begin{equation}
i_*: \pi_1(U, x_0) \to \pi_1(X, x_0) \quad \text{and} \quad j_*: \pi_1(V, x_0) \to \pi_1(X, x_0)
\end{equation}generate $\pi_1(X, x_0)$.

$\textbf{Proof.}$
This theorem states that, given any loop $f$ in $X$ based at $x_0$, it is path homotopic to a product of the form $(g_1 \ast (g_2 \ast(\ldots \ast g_n)))$, where each $g_i$ is a loop in $X$ based at $x_0$ that lies either in $U$ or in $V$.

$\textbf{Step 1.}$ We show there is a subdivision $a_0 < a_1 < \ldots < a_n$ of the unit interval such that $f(a_i) \in U \cap V$ and $f([a_{i-1}, a_i])$ is contained either in $U$ or in $V$ for each $i$.

To begin, choose a subdivision $b_0 , b_1 , \ldots , b_m$ of $[0, 1]$ such that for each $i$, the set $f([b_{i-1}, b_i])$ is contained in either $U$ or $V$ (Use the Lebesgue number lemma). If $f(b_i)$ belongs to $U \cap V$ for each $i$, we are finished. If not, let $i$ be an index such that $f(b_i) \notin U \cup V$. Each of the sets $f([b_{i-1}, b_i])$ and $f([b_i, b_i+1])$ lies either in $U$ or in $V$. If $f(b_i) \in U$, then both of these sets must lie in $U$; and if $f(b_i) \in V$, both of them must lie in $V$. In either case, we may delete $b_i$, obtaining a new subdivision $c_0 , c_1 , \ldots , c_m$ that still satisfies the condition that $f([c_{i-1}, c_i])$ is contained either in $U$ or in $V$, for each $i$. A finite number of repetitions of this process leads to the desired subdivision.

$\textbf{Step 2.}$ We prove the theorem. Given $f$, let $a_0 , a_1 , \ldots , a_n$ be the subdivision constructed in Step 1. Define $f_i$ to be the path in $X$ that equals the positive linear map of $[0,1]$ onto $[a_{i-1}, a_i]$ followed by $f$,Then $f_i$ is a path that lies either in $U$ or in $V$, and
\begin{equation}
[f] = [f_1] \ast [f_2] \ast \ldots \ast [f_m].
\end{equation}

For each $i$, choose a path $\alpha_i$ in $U \cap V$ from $x_0$ to $f(a_i)$ (Here we use the fact that $U \cap V$ is path connected). Since $f(a_0)=f(a_n)=x_0$, we can choose $\alpha_0$ and ${\alpha}_{n}$ to be the constant path at $x_0$. See Figure 59.1.

Now, we set
\begin{equation}
g_i = \left({\alpha}_{n-1} \ast f_i\right) \ast\bar{\alpha}_i
\end{equation}for each $i$.Then $g_i$ is a loop in $X$ based at $x_0$ whose image lies either in $U$ or in $V$. Direct computation shows that
\begin{equation}
[g_1] \ast [g_2] \ast \ldots \ast [g_n] = [f_1] \ast [f_2] \ast \ldots \ast [f_n]
\end{equation}


$\textbf{Corollary}$
Suppose $X = U \cup V$, where $U$ and $V$ are open sets of $X$; suppose $U \cap V$ is nonempty and path connected. If $U$ and $V$ are simply connected, then $X$ is simply connected.

$\textbf{Theorem}$ If $n \geq 2$, the $n$-sphere $S^n$ is simply connected.

$\textbf{Proof.}$
Let $p = (0,\ldots,0, 1)$ and $q = (0, \ldots,0, -1)$ be the "north pole" and the "sousth pole" of $S^n$, respectively.

$\textbf{Step 1.}$ We show that if $n \geq 1$, the punctured sphere $S^n - p$ is homeomorphic to $\mathbb{R}^n$.

Define $f: (S^n - p) \to \mathbb{R}^n$ by the equation
\begin{equation}
f(x) = f(x_1, \ldots,x_{n+1}) = \frac{1}{1 - x_{n+1}}\left(x_1, \ldots, x_n\right).\end{equation}The map $f$ is called $\textit{stereographic projection}$. (If one takes the straight line in $\mathbb{R}^{n+1}$ passing through the north pole $p$ and the point $x$ of $S^n - p$, then this line intersects the $n$-plane $\mathbb{R}^n \times \{0\} \subseteq \mathbb{R}^{n+1}$ in the point $f(x) \times \{0\}$.) One checks that $f$ is a homeomorphism by showing that the map $g: \mathbb{R}^n \to (S^n - p)$ given by
\begin{equation}
g(y) = g(y_1,\ldots,y_n) =(t(y) \cdot y_1, \ldots, t(y) \cdot y_n, 1 - t(y))
\end{equation}
where $t(y) = \frac{2}{1 + \lVert y \rVert^2}$, is a right and left inverse for $f$. Note that the reflection map $(x_1, x_2, \ldots, x_n) \to (x_1, x_2, \ldots, -x_n)$ defines a homeomorphism of $S^n - p$ with $S^n-q$, so the latter is also homeomorphic to $\mathbb{R}^n$.

$\textbf{Step 2.}$ We prove the theorem. Let $U$ and $V$ be the open sets $U = S^n - p$ and $V = S^n-q$ of $S^n$.

Note first that for $n \geq 1$, the sphere $S^n$ is path connected. This follows from the fact that $U$ and $V$ are path connected (being homeomorphic to $\mathbb{R}^n$) and have the point $(1, 0,\ldots ,0)$ of $S^n$ in common.

Now we show that for $n \geq 2$, the sphere $S^n$ is simply connected. The spaces $U$ and $V$ are simply connected, being homeomorphic to $\mathbb{R}^n$. Their intersection equals $S^n -p - q$, which is homeomorphic under stereographic projection to $\mathbb{R}^n - 0$. The latter space is path connected, for every point of $\mathbb{R}^n - 0$ can be joined to a point of $S^{n-1}$ by a straight-line path, and $S^{n-1}$ is path connected if $n \geq 2$. Then the preceding corollary applies.

# Fundamental Groups of Some Surfaces

$\textbf{Theorem}$
The fundamental group of the product space $\pi_1(X \times Y, x_0\times y_0)$ is isomorphic with$\pi_1(X, x_0) \times \pi_1(Y, y_0)$.

$\textbf{Proof.}$
Let $p : X \times Y \to X$ and $q : X \times Y \to Y$ be the projection mappings. Using the base points indicated in the statement of the theorem, we have induced homomorphisms
\begin{equation}
p_* : \pi_1(X \times Y, x_0 \times y_0) \to \pi_1(X, x_0) \\
q_* : \pi_1(X \times Y, x_0 \times y_0) \to \pi_1(Y, y_0).
\end{equation}
We define a homomorphism
\begin{equation}
\Phi : \pi_1(X \times Y, x_0 \times y_0) \to \pi_1(X, x_0) \times \pi_1(Y, y_0)
\end{equation}
by the equation $\Phi([f]) = p_*([f]) \times q_*([f]) = [p \circ f] \times [q \circ f]$. We shall show that $\Phi$ is an isomorphism.

The map $\Phi$ is surjective. Let $g : I \to X$ be a loop based at $x_0$; let $h : I \to Y$ be a loop based at $y_0$. We wish to show that the element $[g] \times [h]$ lies in the image of $\Phi$. Define $f : I \to X \times Y$ by the equation
\begin{equation}
f(s) = g(s)\times h(s).
\end{equation}Then $f$ is a loop in $X \times Y$ based at $(x_0, y_0)$, and
\begin{equation}
\Phi([f]) = [p \circ f] \times [q \circ f] = [g] \times [h],
\end{equation}
as desired.

The kernel of $\Phi$ vanishes. Suppose that $f : I \to X \times Y$ is a loop in $X \times Y$ based at $(x_0\times y_0)$ and $\Phi([f]) = [p \circ f] \times [q \circ f]$ is the identity element. This means that $p \circ f \cong_p e_{x_0}$ and $q \circ f \cong_p e_{y_0}$. Let $G$ and $H$ be the respective path homotopies. Then the map $F : I \times I \to X \times Y$ defined by
\begin{equation}
F(s,t) = G(s,t) \times H(s,t)
\end{equation} is a path homotopy between $f$ and the constant loop based at $x_0\times y_0$.

$\textbf{Corollary}$
The fundamental group of the torus $T = S^1 \times S^1$ is isomorphic to the group $\mathbb{Z} \times \mathbb{Z}$.

$\textbf{Definition.}$
The projective plane $P^2$ is the quotient space obtained from $S^2$ by identifying each point $x$ of $S^2$ with its antipodal point $-x$.

$\textbf{Theorem}$
The projective plane $P^2$ is a compact surface, and the quotient map $p: S^2 \to P^2$ is a covering map.

$\textbf{Proof}$
First, we show that $p$ is an open map. Let $U$ be open in $S^2$.Now the antipodal map $a: S^2 \to S^2$ given by $a(x) = -x$ is a homeomorphism of $S^2$. Hence, $a(U)$ is open in $S^2$. Since
\begin{equation}
p^{-1}(p(U)) = U \cup a(U),
\end{equation}this set is also open in $S^2$. Therefore, by definition, $p(U)$ is open in $P^2$. A similar proof shows that $p$ is a closed map.

Now, we show that $p$ is a covering map. Given a point $y$ of $P^2$, choose $x \in p^{-1}(y)$. Then choose an $\varepsilon$-neighborhood $U$ of $x$ in $S^2$ for some $\varepsilon < 1$, using the Euclidean metric $d$ of $\mathbb{R}^3$. Then $U$ contains no pair $\{z, a(z)\}$ of antipodal points of $S^2$ since $d(z, a(z)) = 2$. As a result, the map
\begin{equation}
p: U \to p(U)
\end{equation}is bijective. Being continuous and open, it is a homeomorphism. Similarly,
\begin{equation}
p : a(U) \to p(a(U)) = p(U)
\end{equation}is a homeomorphism. The set $p^{-1}(p(U))$ is thus the union of the two disjoint open sets $U$ and $a(U)$, each of which is mapped homeomorphically by $p$ onto $p(U)$. Then $p(U)$ is a neighborhood of $p(x) = y$ that is evenly covered by $p$.

==Since $S^2$ has a countable basis $\{U_n\}$, the space $P^2$ also has a countable basis $\{p(U_n)\}$.==

The fact that $P^2$ is ==Hausdorff== follows from the fact that $S^2$ is normal and $p$ is a closed map. Alternatively, one can give a direct proof: Let $y_1$ and $y_2$ be two points of $P^2$. The set $p^{-1}(y_1) \cup p^{-1}(y_2)$ consists of four points; let $2\varepsilon$ be the minimum distance between them. Let $U_{1}$ be the $\varepsilon$-neighborhood of one of the points of $p^{-1}(y_1)$ and let $U_{2}$ be the $\varepsilon$-neighborhood of one of the points of $p^{-1}(y_2)$. Then
\begin{equation}
U_{1} \cup a(U_1) \quad and \quad U_{2} \cup a(U_2)
\end{equation}are disjoint. It follows that $p(U_{1})$ and $p(U_{2})$ are disjoint neighborhoods of $y_1$ and $y_2$ respectively, in $P^2$.

Since $S^2$ is a surface and every point of $P^2$ has a neighborhood homeomorphic with an open subset of $S^2$, the space $P^2$ is also a surface.

$\textbf{Corollary}$
$\pi_1(P^2, y)$ is a group of order $2$.

$\textbf{Proof}$
The projection $p: S^2 \to P^2$ is a covering map. Since $S^2$ is simply connected, we can apply Theorem 54.4, which tells us there is a bijective correspondence between $\pi_1(P^2, y)$ and the set $p^{-1}(y)$.Since this set is a two-element set, so $\pi_1(P^2, y)$ is a group of order 2. Any group of order 2 is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, the integers modulo 2.

$\textbf{Theorem 54.4}$
Let $p: E \rightarrow B$ be a covering map, and let $p(e_0) = b_0$. If $E$ is path connected, then the lifting correspondence $\phi: \pi_1(B, b_0) \rightarrow p^{-1}(b_0)$ is surjective. If $E$ is simply connected, it is bijective.

$\textbf{Lemma}$
The fundamental group of the figure eight is not abelian.

$\textbf{Proof}$
Let $X$ be the union of two circles $A$ and $B$ in $\mathbb{R}^2$ whose intersection consists of the single point $x_0$. We describe a certain covering space $E$ of $X$.

The space $E$ is the subspace of the plane consisting of the x-axis and the y-axis, along with tiny circles tangent to these axes, one circle tangent to the x-axis at each nonzero integer point and one circle tangent to the y-axis at each nonzero integer point.

The projection map $p : E \rightarrow X$ wraps the x-axis around the circle $A$ and wraps the y-axis around the other circle $B$; in each case, the integer points are mapped by $p$ into the base point $x_0$. Each circle tangent to an integer point on the x-axis is mapped homeomorphically by $p$ onto $B$, while each circle tangent to an integer point on the y-axis is mapped homeomorphically onto $A$; in each case, the point of tangency is mapped onto the point $x_0$. It's easy to see that the map $p$ is indeed a covering map.

Now let $\tilde f : I \rightarrow E$ be the path $\tilde f(s) = s \times 0$, going along the x-axis from the origin to the point $1 \times 0$. Let $g : I \rightarrow E$ be the path $\tilde g(s) = 0 \times s$, going along the y-axis from the origin to the point $0 \times 1$. Let $f = p \circ \tilde f$ and $g = p \circ \tilde g$; then $f$ and $g$ are loops in the figure eight based at $x_0$, going around the circles $A$ and $B$ respectively. See Figure 60.1.


We assert that $f * g$ and $g * f$ are not path homotopic, so that the fundamental group of the figure eight is not abelian.

To prove this assertion, let us lift each of these to a path in $E$ beginning at the origin. The path $f * g$ lifts to a path that goes along the x-axis from the origin to $1 \times 0$ and then goes once around the circle tangent to the x-axis at $1 \times 0$. On the other hand, the path $g * f$ lifts to a path in $E$ that goes along the y-axis from the origin to $0 \times 1$, and then goes once around the circle tangent to the y-axis at $0 \times 1$. Since the lifted paths do not end at the same point, $f * g$ and $g * f$ cannot be path homotopic.

$\textbf{Theorem}$
The fundamental group of the double torus is not abelian.

$\textbf{Proof.}$
The double torus $T\#T$ is the surface obtained by taking two copies of the torus, deleting a small open disc from each of them, and pasting the remaining pieces together along their edges. We assert that the figure eight $X$ is a retract of $T\#T$. This fact implies that the inclusion $j: X \to T\#T$ induces a monomorphism, so that $\pi_1(X, x_0)$ is not abelian.

One can write equations for the retraction $r: T\#T \to X$, but it is simpler to indicate it in pictures, as we have done in Figure 60.2. Let $Y$ be the union of two tori having a point in common. First, one maps $T\#T$ onto $Y$ by a map that collapses the dotted circle to a point but is otherwise one-to-one; it defines a homeomorphism $h$ of the figure eight in $T\#T$ with the figure eight in $Y$. Then one retracts $Y$ onto its figure eight by mapping each cross-sectional circle to the point where it intersects the figure eight. Then one maps the figure eight in $Y$ back onto the figure eight in $T\#T$ by the map $h^{-1}$.


$\textbf{Corollary}$
The torus, projective plane, and double torus are topologically distinct.