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6.4.2 isn't using funext, close #640 #643

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Mar 9, 2014
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6.4.2 isn't using funext, close #640 #643

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ec1373a
6.4.2 isn't using funext, close #640
mikeshulman Mar 2, 2014
be2d65a
errata for ec1373a
mikeshulman Mar 2, 2014
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4 changes: 4 additions & 0 deletions errata.tex
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Expand Up @@ -480,6 +480,10 @@
& 417-g4aa6a15
& Added \autoref{ex:funext-from-interval}: the function constructed in \autoref{thm:interval-funext} is actually an inverse to $\happly$, so that the full function extensionality axiom follows from an interval type.\\
%
\autoref{thm:S1-autohtpy}
& % merge of ec1373a
& In the second paragraph of the proof, the appeal to function extensionality should be omitted.\\
%
\autoref{sec:circle}
& 327-g7cbe31c
& In the first sentence after the proof of \autoref{thm:apd2}, ``$P:\Sn^2\to P$'' should be ``$P:\Sn^2\to\type$''.\\
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2 changes: 1 addition & 1 deletion hits.tex
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Expand Up @@ -459,7 +459,7 @@ \section{Circles and spheres}
However, in \autoref{sec:compute-paths} we observed that $\transfib{x\mapsto x=x}{p}{q} = \opp{p} \ct q \ct p$, so what we have to show is that $\opp{\lloop} \ct \lloop \ct \lloop = \lloop$.
But this is clear by canceling an inverse.

To show that $f\neq (x\mapsto \refl{x})$, it suffices by function extensionality to show that $f(\base) \neq \refl{\base}$.
To show that $f\neq (x\mapsto \refl{x})$, it suffices to show that $f(\base) \neq \refl{\base}$.
But $f(\base)=\lloop$, so this is just the previous lemma.
\end{proof}

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