Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let us consider the following bijective applications: $$f_{k}:A_{k}→B_{k}, k=1,2,\ldots,m$$ We construct the following application: $$f=(f_1,f_2,\ldots,f_m):A_1×A_2\times\cdots ×A_m→B_1×B_2\times\cdots \times B_m$$

Hence $f$ is a vector-valued function from $A_{1}×A_{2}\times\cdots ×A_{m}$ to $B_{1}×B_{2}\times \cdots ×B_{m}$. If the determinant of the Jacobian matrix is nonzero everywhere, then by the Inverse function theorem, for every point $p$ in $A_{1}×A_{2}\times\cdots×A_{m}$, there exists a neighborhood about $p$ over which $f$ is invertible. My question is: Find sufficient and necessary conditions in which $f$ is invertible over its entire image.

share|improve this question
1  
If $f_i$ is bijective and depends only on the $i$th component of the input, isn't $f$ always a bijection? –  user7530 Sep 13 '13 at 14:00
1  
It seems to me that $f$ is necessarily bijective with $f^{-1}: B_1 \times B_2 \times ... \times B_M \to A_1 \times A_2 \times ... \times A_M \\(b_1, b_2, ..., b_m) \mapsto (f_1 ^{-1}(b_1), f_2 ^{-1}(b_2), ..., f_m ^{-1}(b_m))$ –  Daron Sep 13 '13 at 14:04
1  
@Ar1 a bijection absolutely doesn't need to be continuous or even differentiable –  Dominic Michaelis Sep 13 '13 at 14:08
1  
@AR1 ok then an injection absolutely doesn't need to be differentiable –  Dominic Michaelis Sep 13 '13 at 14:14
1  
@Ar1 I feel like I am repeating. Point 1: let $f: X\to Y$ be any function. Then $\tilde{f}:X\to f(X)$ is a surjection. So if $f$ is injective then $\tilde{f}:X\to f(X)$ is a bijection. –  Dominic Michaelis Sep 13 '13 at 14:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.