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Why is the identity map from $S^1$ to $S^1$ not homotpic to the constant map?

I get the picture that if the identity map $id$ is homotopic to the constant map then as the circle transforms through the homotopy it would break. But I cannot come up with a formal proof. Can someone offer a proof without using "contractable"? Here's what we have covered during the class:

  1. Two maps being homotopic to each other
  2. Two space being homotopy equivalent
  3. Quotient space
  4. One-point compactification
  5. What a covering map is, and what path-lifting property is.

Thank you very much.

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    $\begingroup$ No, we can't. You're asking us to show that the fundamental group of $S^1$ is nontrivial, and to do this, you're going to need to build up some sort of machinery. One common bit is to talk a little bit about "covering maps" and the "homotopy lifting property" of the map $\Bbb R \to S^1$, $t \mapsto e^{it}$. You don't necessarily need this particular bit of machinery to show that $\pi_1(S^1)$ is nontrivial, but you need something. $\endgroup$ – user98602 May 22 '15 at 22:32
  • $\begingroup$ @MikeMiller We did briefly talked about "path-lifting property" as well as what a covering map is. $\endgroup$ – 3x89g2 May 22 '15 at 22:36
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    $\begingroup$ It's usually very difficult to prove directly that two spaces are not homotopy equivalent, homeomorphic, etc. Instead, the usual approach is to consider some sort of algebraic structure, show that it's invariant under homotopy equivalence, homeomorphism, etc., and then use that to distingush the two spaces. If you've only covered those four things in your class so far, then frankly you don't have enough machinery to handle this problem easily. (Also, a contractible space is exactly one in which the identity map is null-homotopic, so I'm not sure how to avoid talking about that concept.) $\endgroup$ – anomaly May 22 '15 at 22:37
  • $\begingroup$ @anomaly I see. Our professor said he proved that proposition during the lecture and unfortunately I missed that one. I tried to ask my classmates and apparently no one got what he said :( $\endgroup$ – 3x89g2 May 22 '15 at 22:38
  • $\begingroup$ @anomaly And now we are "proving" that the identity of $S^n$ is not homotopy equivalent to $S^m$. I guess I'll have to take that for granted. $\endgroup$ – 3x89g2 May 22 '15 at 22:39
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Embedding in $\mathbb C$, the identity is $z\mapsto z^1$ and the constant map is $z\mapsto 1=z^0$. If $(z\mapsto z^n)\simeq(z\mapsto z^m)$ let $\pi_t:S^1\times I\to S^1$ be such a homotopy with $$\pi_0=(z\mapsto z^n),\pi_1=(z\mapsto z^m).$$ Lift to a homotopy $\bar\pi_t$ of paths in $\mathbb R$ starting at $0$, noting $\bar\pi_0=\bar p_n$ and $\bar\pi_1=\bar p_m$ where $p_i$ is $z\mapsto z^i$. Then $\bar\pi_t(1)$ is independent of $t$, so $$n=\bar\pi_0(1) = \bar\pi_1(1)=m.$$

"Now use $0\neq 1$." This answer uses only elementary properties of covering spaces.

Note: in the last equation, the first and third equalities come from the fact that $n$th power mappings in $S^1$ lift to $s\mapsto ns$ in $\mathbb R$.

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This assume that the homotopy can be defined by a smooth map.

Consider $\mathbb S^1 \subset \mathbb R^2$. Then $\gamma : \mathbb S^1 \to \mathbb S^1$ can be treated as a curve in $\mathbb R^2$.

Now calculate $\int_\gamma d\theta$, where

$$d\theta = \frac{1}{x^2 + y^2} (-y\,dx + x\,dy).$$

When $\gamma = id$, this is $2\pi$. But for homotopic maps, the value should be the same (by Stokes theorem). Thus the identity is not homotopic to a constant map.

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$\newcommand{\Reals}{\mathbf{R}}$Here's a sketch using results you know about covering spaces and lifts. Express the circle $S^{1}$ as $\Reals/(2\pi\mathbf{Z})$. The idea is, if the identity map of the circle were homotopic to a constant map, then the identity map of the reals would be homotopic to a constant map through maps sending the "lattice" $2\pi\mathbf{Z}$ to itself.

For definiteness, let $p:\Reals \to S^{1}$ be the natural quotient map (a.k.a. the universal covering map), $id:S^{1} \to S^{1}$ the identity map, and $c:S^{1} \to S^{1}$ the constant map whose value is the coset $[0] = 2\pi\mathbf{Z}$.

If $h_{0}:S^{1} \times [0, 1] \to S^{1}$ were a homotopy from $id$ to $c$, then the mapping $h:\Reals \times [0, 1] \to S^{1}$ defined by $h(x, t) = h_{0}\bigl(p(x), t\bigr)$ would be a homotopy between $p = h(\cdot, 0)$ and $c \circ p = h(\cdot, 1)$. By the lifting property there would exist a homotopy $H:\Reals \times [0, 1] \to \Reals$ such that $$ H(x, 0) = x,\quad H(x, 1) = 0\qquad\text{for all real $x$,} $$ and $H(2\pi, t) \in p^{-1}([0]) = 2\pi \mathbf{Z}$ for all $t$ in $[0, 1]$ (because $p \circ H = h$). That's obviously impossible. (The intermediate value theorem can be used to give a rigorous proof.)

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  • $\begingroup$ I refer to this mathoverflow discussion mathoverflow.net/questions/40945/…, and the picture in my answer. The point is that the result asked for is quite subtle, but should be seen as part of the more general situation of gluing non path connected spaces. A tool here is the notion of the fundamental groupoid on a set of base points. $\endgroup$ – Ronnie Brown May 23 '15 at 9:13

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