Building the inverse

Question

If $\theta_1$ and $\theta_2$ are dependent, let $a_2$ be fixed at $\bar{a}_2$

$x_2= x_2(\bar{a}_2,\theta_2) \rightarrow \theta_2 =x_2^{-1}(\bar{a}_2,x_2)$

Why can you build the inverse dependend on $x_2$?

Thank you :)

Here is the inverse in the proof i try to understand

Answer 1

If a function $g$ has an inverse $g^{-1}$, then by definition: $$g^{-1}(g(z))=z.$$

The proof refers to the inverse of $x_1$ with respect to its second coordinate $\theta_1$. The inverse of $x_1$ with respect to $\theta_1$ exists because $x_1$ is (strictly) monotone in $\theta_1$. The inverse $x_1^{-1}$ of $x_1$ must satisfy $$x_1^{-1}(a_1,x_1(a_1,\theta_1))=\theta_1.$$ In the linked image, the argument $(a_1,\theta_1)$ of $x_1$ has been suppressed:

$$x_1^{-1}(a_1,x_1)=\theta_1.$$