How is $\mathbb R \to S^1$ nullhomotopic?

Question

Every map $\mathbb R \to S^1$ is nullhomotopic.

The covering map $p: \mathbb R \to S^1$ is given by $p(x)=\cos(2\pi x, \sin 2\pi x)$.

How is this map null-homotopic if we are wrapping $\mathbb R$ infinitely many times around the circle?

Answer 1

Suppose $X$ is contractible and let $f: X \to Y$ be any continuous map into any space $Y$. Since $X$ is contractible then $Id_X \simeq c_{x_0}$ where $c_{x_0}: X \to X$ is the map sending everything to $x_0$. So, $$f=f \circ Id_X \simeq f \circ c_{x_0} = c_{f(x_0)}.$$ So, $f$ is null-homotopic.

Conversely, if any $f: X \to Y$ is null-homotopic for any $Y$, then let $Y=X$ and let $f=Id_X$. Then, $Id_X$ is null-homotopic and thus $X$ is contractible.

---

We can also show $Y$ is contractible iff every $f: X \to Y$ is null-homotopic for any $X$.

Answer 2

For a fixed parameter $t$ consider the map $\mathbb{R} \to S^1$ given by $$ p_t(x) = (\cos(2\pi t x), \sin(2 \pi t x)). $$ Then $p_1$ is your map, and $p_0$ is the constant map sending everything to $(1, 0)$.

Answer 3

Hint: $\Bbb R$ is contractible.

Answer 4

Lets's prove something more general. Take $X$ a contractible space with a homotopy equivalence $H : 1_X \simeq c_{x_0}$, and let $f : X \to Y$ be a continuous map. Now, the function

$$ \begin{align} L : \ &I \times X \to Y \\ & (t,x) \mapsto f(H(t,x)) \end{align} $$

is continuous, and $L_0(x) = f(H_0(x)) = f(x)$, $L_1(x) = f(H_1(x)) = f(x_0)$. Hence $f$ is homotopy equivalent to the constant function $f(x_0)$. We have thus proven:

Proposition. Let $X$ be a contractible space. Then, any continuous function $f : X \to Y$ is nullhomotopic.

And of course, since $\mathbb{R}$ is contractible, this solves your problem.

Even if we map into the circle making an infinite amount of 'wrapping', any path $[0,1] \to \mathbb{R}$ we choose maps to a path which we are able to contract. This is weaker than being able to do the same with the whole line, which (I think) is were your misconception stems from. Also, note that this has to do with the flexibility of contractions being able to move all points (which is not allowed when talking about elements of a certain fundamental group).