Ramana Kumar comments on Committing, Assuming, Externalizing, and Internalizing

[Ramana Kumar]

$h_i^{\prime \prime }:A_{1-i}/B_{1-i}\times E_{1-i}\to A_i/B_i\times E_i$ is given by $h_i^{\prime }(b,e)=\left(f_{1-i}\right(b),h_i(e\left)\right)$

There’s a prime missing on $h_i^{\prime }$. I’d also have expected $q$ instead of $b$ as the variable (doesn’t affect correctness).

[Scott Garrabrant]

Fixed, thanks.