# Determinant of function composed with its inverse

Let $F:A \to B$ map subsets of $\mathbb{R}^n$ with inverse $F^{-1}$.

Let $d(\cdot) = \text{det} \mathbf{D}F(\cdot)$ with $\mathbf{D}$ denoting the total derivative matrix.

Am I correct that $d \circ F^{-1} = \text{det} \mathbf{D}F(F^{-1}) = \text{det} \mathbf{D}\text{(Id)} = 1$.

I am not entirely sure of the what the definition of $d$ really means so I better ask this question. Thanks.

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For each $a\in A$, $\det(\mathbf{D}F(a))$ is the determinant of the matrix $\mathbf{D}F(a)$. For me it does not make sense to evaluate $\mathbf{D}F(F^{-1})$. – Sigur Dec 23 '12 at 17:40

No. The function $d\circ F^{-1}$ is $\mathbf DF\circ F^{-1}$, not $\mathbf D(F\circ F^{-1})$.
Thanks. Do you know any way to simplify $d \circ F^{-1}$. – Lemon Dec 23 '12 at 18:29