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


*

*Two maps being homotopic to each other

*Two space being homotopy equivalent

*Quotient space

*One-point compactification

*What a covering map is, and what path-lifting property is.


Thank you very much.
 A: 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 $\mathbb R$.
A: 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.
A: $\newcommand{\Reals}{\mathbf{R}}$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.)
