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Suppose $\det f^\prime(a) \neq 0$, $f$ is continuously differentiable in an open set containing $a$. Spivak then show that there exist a closed rectangle $U$ containing $a$ in its interior such that $f(x) \neq f(a)$ for all $x \in U$.

Question1: Spivak then asserts that since $f$ is continuously differentiable in an open set containing $a$, therefore $\det f'(x) \neq 0$ for all $x \in U$. Why?

Question 2: $|D_j f^i (x)-D_j f^i (a)|<\tfrac{1}{2n^2}.$ How can we prove it?

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  1. Since $f$ is continuously differentiable on a neighbourhood $V$ of $a$, $x \mapsto \det f^\prime(x)$ is continuous on $V$, so that since $\det f^\prime(a) \neq 0$, one can choose a sufficiently small closed rectangle $U_0$ containing $a$ as an interior point, such that $\det f^\prime(x) \neq 0$ for $x \in U_0$.

  2. Again, since $f$ is continuously differentiable on a neighbourhood $V$ of $a$, $x \mapsto f^\prime(x)$ is continuous on $V$, so one can choose a sufficiently small closed rectangle $U_1 \subseteq U_0$ containing $a$ as an interior point, such that $\| f^\prime(x) - f^\prime(a)\| < \tfrac{1}{2n^2}$ for $x \in U_1$ for $\|\cdot\|$ the norm on the relevant vector space of real matrices; in particular, then, for each $i$ and $j$, $$|D_j f^i(x) - D_j f^i(a)| \leq \| f^\prime(x) - f^\prime(a)\| < \tfrac{1}{2n^2}.$$

Since $U_1 \subseteq U_0$, $U = U_1$ does the trick, at least for claims 1. and 2.

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  • $\begingroup$ If f is continuously differentiable then how can we prove that x→det f'(x) is continuous $\endgroup$ – Tauseef Khan Jan 5 '13 at 14:26
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    $\begingroup$ Since $f : V \subset \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable, $f^\prime : V \to M_n(\mathbb{R})$ is continuous; on the other hand, $\det : M_n(\mathbb{R}) \to \mathbb{R}$ is continuous, as the determinant of a matrix is a polynomial in its entries. Thus, $x \mapsto \det f^\prime(x)$ is a composition of continuous functions, and hence itself continuous. $\endgroup$ – Branimir Ćaćić Jan 5 '13 at 15:01

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