TITLE: Exercise 4.6 (iii) in the HoTT Book
QUESTION [9 upvotes]: Being unable to solve Exercise 4.6 (iii) in the HoTT Book, I'd be most grateful for any help.
Recall the statement of Exercise 4.6:
For $A,B:\mathcal U$, define
$$
\mathsf{idtoqinv}_{A,B}:(A=B)\to\sum_{f:A\to B}\mathsf{qinv}(f)
$$
by path induction in the obvious way. Let $\mathsf{qinv}$-$\mathsf{univalence}$ denote the modified form of the univalence axiom which asserts that for all $A,B:\mathcal U$ the function $\mathsf{idtoqinv}_{A,B}$ has a quasi-inverse.
(i) Show that $\mathsf{qinv}$-$\mathsf{univalence}$ can be used instead of univalence in the proof of function extensionality in $\S$4.9.
(ii) Show that $\mathsf{qinv}$-$\mathsf{univalence}$ can be used instead of univalence in the proof of Theorem 4.1.3.
(iii) Show that $\mathsf{qinv}$-$\mathsf{univalence}$ is inconsistent (i.e. allows construction of an inhabitant of 0). Thus, the use of a "good" version of $\mathsf{isequiv}$ is essential in the statement of univalence.

REPLY [4 votes]: As observed by Luis, it suffices to show that, for any map $f:A\to B$, the type $\mathsf{qinv}(f)$ is a mere proposition. 
For any types $A$ and $B$ we put 
$$
(A\simeq B):\equiv\sum_{f:A\to B}\mathsf{ishae}(f),\quad(A\backsimeq B):\equiv\sum_{f:A\to B}\mathsf{qinv}(f).
$$ 
(Recall that $\mathsf{ishae}(f)$ stands for "$f$ is a half-adjoint equivalence".) By Theorem 4.2.3 and the comment preceding it, there are dependent functions 
$$
i:\prod_{A,B:\mathcal U}\prod_{f:A\to B}\mathsf{ishae}(f)\to\mathsf{qinv}(f),\quad p:\prod_{A,B:\mathcal U}\prod_{f:A\to B}\mathsf{qinv}(f)\to\mathsf{ishae}(f).
$$ 
For all $A,B:\mathcal U$ let $e(A,B):(A=B)\to(A\backsimeq B)$ be the natural map (which we assume to admit a quasi-inverse).
Claim: if $f:A\to B$ and $x:\mathsf{qinv}(f)$, then $i(A,B,f,p(A,B,f,x))=x$.
Since $(f,x):A\backsimeq B$, we can assume by $\mathsf{qinv}$-univalence and path induction that 
$$
B\equiv A,\quad f\equiv\mathsf{id}_A,\quad x\equiv e(A,A,\mathsf{refl}_A),
$$ 
and the claim follows easily from the definition of $i,p$ and $e$.
For $f:A\to B$ and $x,y:\mathsf{qinv}(f)$ we have $p(A,B,f,x)=p(A,B,f,y)$ because $\mathsf{ishae}(f)$ is a mere proposition, and the claim implies $x=y$, as required.