Consider a map $s\in C^1(Q,\Omega)$ where $Q,\Omega\subset\mathbb{R}^2$ are bounded and have Lipschitz boundary. Moreover $s$ is invertible and $||s'||_{L^{\infty}}<c$ ,$||(s')^{-1}||_{L^{\infty}}<c$, $det (s')> 0$.
Is it possible to obtain a lower bound for $det(s')$ from the upper bound of $(s')^{-1}$?
The answer is yes. For a linear map $A$ over $\Bbb R^2$, we have $$ |\det(A)| = \sigma_{\max}(A) \sigma_{\min}(A) = \frac{\|A\|_{L^2}}{\|A^{-1}\|_{L^{2}}}, $$ where $\sigma$ refers to a singular value of $A$. We also have $\|A\| \geq \frac{1}{\|A^{-1}\|}$. Noting that $\frac 1{\sqrt n} \|A\|_{L^\infty}\leq \|A\|_{L^2} \leq \sqrt{n} \|A\|_{L^\infty}$, we can conclude that $$ |\det(s')| = \frac{\|s'\|_{L^2}}{\|(s')^{-1}\|_{L^2}} \geq \frac{1}{\|(s')^{-1}\|_{L^2}^2} \geq \frac{1}{n\|(s')^{-1}\|_{L^\infty}^2} > \frac 1{n c^2} = \frac 1{2c^2}. $$