You can't have a function which, for all A, gives you ||A|| -> A. See the exercises 3.11 and 3.12!
-- Nicolai

On 05/03/19 22:31, Jean Joseph wrote:
Hi,

From the HoTT book, the truncation of any type A has two constructors:

1) for any a : A, there is |a| : ||A||
2) for any x,y : ||A||, x = y. 

I get that if A is inhabited, then ||A|| is inhabited by (1). But is it true that, if ||A|| is inhabited, then A is inhabited? 
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