(p : (x : A) -> Id (B x) (f x) (g x)) : Id ((y : A) -> B y) f g =
<i> \(a : A) -> (p a) @ i
--- To see the definition of funExt makes sense we can compute its faces:
---
--- funExt @ 0 = \(a : A) -> (p a) @ 0 (faces commute with operations)
--- = \(a : A) -> f a (p a proves Id (B a) (f a) (g a))
--- = f (by eta)
+{-
+
+To see that the definition of funExt makes sense we can compute its
+faces:
+
+funExt @ 0 = \(a : A) -> (p a) @ 0
+ = \(a : A) -> f a
+ = f
+
+The last equality holds because of eta.
+
+-}
{-