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Nov 5, 2013 at 12:03 comment added Urs Schreiber Sorry about the "free", should have said "presentable". Fixed now.
Nov 5, 2013 at 12:03 history edited Urs Schreiber CC BY-SA 3.0
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Nov 5, 2013 at 6:08 comment added Mike Shulman No, it's not in principle straightforward. There are coherence issues. On the other hand, HITs don't just present free $\infty$-monads; indeed the capability to present non-free monads is more or less what distinguishes HITs from ordinary inductive types.
Nov 4, 2013 at 18:53 history answered Urs Schreiber CC BY-SA 3.0