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Why isn't $S^1$ contractible in homotopy type theory?

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 ...
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Why isn't $S^1$ contractible in homotopy type theory?

The question was answered by Noah Snyder in the comments: No, $S^1$ is not contractible. It connected, so any point in $S^1$ is merely equal to $base$. To prove that $S^1$ is not contractible, you ...

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