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I have a question which asks me to show that the map f: $R^2 \rightarrow R^2$ defined by $$f(x,y)=(e^x+e^y, e^x+e^{-y})$$ is locally invertible about any point $(a,b) \in R^2$, and compute the Jacobian matrix of the inverse map.

I know if is locally invertible because the determinant of the Jacobian matrix is not zero. However, how do I find the Jacobian matrix of the inverse map? Do I find the inverse matrix directly or should I find the inverse of this map and then find the Jacobian?

There are several similar problems so I would really appreciate it if someone could walk me through the process.

Isa
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Suppose $f \colon \mathbb{R}^n \to \mathbb{R}^n$ is a differentiable function locally invertible at the origin, and let $\operatorname{id} \colon \mathbb{R}^n \to \mathbb{R}^n$ denote the identity map sending every point to itself. Then, by the chain rule, we have that $\operatorname{Jac}(\operatorname{id})|_0 = \operatorname{Jac}(f^{-1} \circ f)|_0 = \operatorname{Jac}(f^{-1})|_{f(0)} \cdot \operatorname{Jac}(f)|_0$. But $\operatorname{Jac}(\operatorname{id})|_0$ is just the identity matrix, so $\operatorname{Jac}(f^{-1})|_{f(0)} = (\operatorname{Jac}(f)|_0)^{-1}$.

The above argument implies that you can find the inverse matrix, or take the Jacobian of the inverse map, and you'll get the same result either way.

Ashvin Swaminathan
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