Why $S^1$ Isn’t Contractible in Homotopy Type Theory

homotopy-type-theory

$\newcommand\base{\mathit{base}}\newcommand\unique{\mathit{unique}}\DeclareMathOperator\transport{transport}\newcommand\loop{\mathit{loop}}\DeclareMathOperator\refl{refl}$In the context of homotopy type theory, I am trying to prove that the only element of $S^1$ is $\base$ i.e. $\unique: \prod_{x:S^1} \base=x$ however by using the induction principle of the circle I reach a dead end:

$$P: S^1 \rightarrow \mathcal{U}; P \mathrel{:\equiv} x \Rightarrow \base=x.$$

For induction, it is required an element $a: P(\base)$ and a path $l: \transport^P(\loop, a) = a$ that by transport over path spaces results in $l: a \cdot \loop = a$.

Choosing $a \mathrel{:\equiv} \refl_{\base}$ or $a \mathrel{:\equiv} \loop$ results on impossible paths (assuming univalence) i.e. $l: \loop = \refl_{\base}$ and $l: \loop \cdot \loop = \loop$.

This made me think if $S^1$ have other elements other than base? or there is some other path $a: \base = \base$ that makes $l$ be derivable, but which ones?

Best Answer

Perhaps this will help: in homotopy type theory, to say that $x$ and $y$ of type $A$ are different corresponds intuitively to the fact that they are in different connected components, i.e., there is no path between them.

Specifically, for the circle $S^1$:

  • The statement $\Pi(x, y: S^1).\, x = y$ ("All points are equal") should be understood geometrically as saying that there is a continuous map which to each pair of points assigns a path between them.

  • The statement $\Sigma(x, y : S^1).\, x \neq y$ ("There are two different points") should be understood as saying that there are points on the circle which are not connected by a path.

Neither of the above is the case. What is the case, though is that $\Pi (x, y : S^1) . \|x = y\|_{-1}$ ("Every two points are in the same connectedness component.")