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Proposition $\pi_1(S^n)=0$ if $n \geq 2$.

Let $f$ be a loop in $S^n$ at a chosen basepoint $x_0$. If the image of $f$ is disjoint from some other point $x \in S^n$ then f is nullhomotopic since $S^n - \{x\}$ is homeomorphic to $\mathbb{R^n}$, which is simply connected.

This is from Hatcher's algebraic topology. The second sentence seems to come out of nowhere. Does anyone have another explanation as to what the second sentence means and is aiming for?

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  • $\begingroup$ Unless the loop $f$ is "spacefiliing" (it occupies all of $S^n$) then it is contained in a subspace homeomorphic to $\Bbb R^n$, and so is contractible. $\endgroup$ Feb 17, 2019 at 10:48
  • $\begingroup$ Another way of looking at that line is that if you're missing a point, then you can perform a stereographic projection to $\mathbb{R}^n$ from that point. $\endgroup$ Feb 17, 2019 at 12:27

3 Answers 3

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That second sentence does not come out of nowhere, it comes out of a case analysis: given a loop $f : [0,1] \to S^n$, $f(0)=f(1)=x_0$, either $f$ is surjective or nonsurjective.

That sentence handles the case where $f$ is nonsurjective.

Now it's time to handle the other case: suppose $f$ is surjective...

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Let $f$ be a loop in $S^n$ based at a point $x_0$ and assume that another point $x\in S^n$ does not lie in the image of $f$. Now choose a homeomorphism $\varphi:S^n\setminus\{x\}\cong\mathbb{R}^n$. Then $\varphi\circ f$ is a loop in $\mathbb{R}^n$ based at the point $\varphi(x_0)$, and since $\mathbb{R}^n$ is simply connected we can choose a null-homotopy $F:\varphi\circ f\sim \ast$ of this loop which is constant on $\ast\times I$. Then $\varphi^{-1}\circ F$ is a null-homotopy of $f=\varphi^{-1}\circ\varphi\circ f$ to the loop which is constant at $x_0$, and is again stationary on $\ast\times I$.

Technically $\varphi^{-1}\circ F$ is a map $S^1\times I\rightarrow S^n\setminus\{x_0\}$, but its composite with the inclusion $S^n\setminus\{x_0\}\subseteq S^n$ is still continuous and is exactly a null-homotopy of $f$ relative to $x_0$. Hence Hatcher's statement.

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Given any point $x_0 \in \mathbb{S}^n$, there is a homeomorphism $\mathbb{S}^n \setminus \{ x_0 \} \cong \mathbb{R}^n$. This is given by stereographic projection. For more on this look here. So therefore, $$\pi_1(\mathbb{S}^n) \cong \pi_1(\mathbb{R}^n) = 0.$$

The first part of that second sentence is just saying that you should choose that point $x_0$ to be any point that isn't on the loop descibed by $f$.

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