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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a vector-field where the function $f_i$ for each dimension $i$ is continuous and strictly increasing. Are these assumptions enough to conclude that the vector-field $f$ is surjective and injective?

As I can understand from the comments, the term "strictly increasing" is not well-defined for vector-fields. What I mean is that if $\vec{x}$ and $\vec{y}$ are two vectors, then I say that $\vec{x} < \vec{y}$ if $\exists i: x_i < y_i \wedge \forall j \neq i: x_j \leq y_j$. So by strictly increasing $f_i$ I mean that if $\vec{x} < \vec{y}$ then $f_i(\vec{x}) < f_i(\vec{y})$.

questiondude
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    What do you mean by "strictly increasing" for a vector field. As far as I know, this term is not defined. As to your problem: are there restrictions for $\vec x$, too? (Like, e.g., elements in $(0,1)$) – amsmath Sep 04 '21 at 16:20
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    This is not easy. First, the entries of $\vec x$ should be non-negative. Otherwise, the square root might not be defined. For $n=2$ there indeed exists a solution, given by $x_j = \sqrt{a_j^2 - z}$, where $z = \sqrt{2\gamma a_1^2a_2^2 + \gamma^2(a_1^2+a_2^2)^2} - \gamma(a_1^2+a_2^2)$ and $\gamma = \frac{c^2}{2(1-c^2)}$ with $c$ being the off-diagonal element of $C$. – amsmath Sep 04 '21 at 18:49
  • Thanks very much for the quick answer and taking a crack at solving this! I have clarified my question a bit. By strictly increasing I mean the individual functions in each dimension $f_i$ are strictly increasing. Intuitively I would think that if each $f_i$ is continuous and strictly increasing, then the entire vector-field $f$ would be bijective. But I'm not sure this actually holds in general, because multi-dimensional functions can behave in weird ways. – questiondude Sep 05 '21 at 10:18
  • I have also added a link to another post where I give an algorithm for solving this problem numerically, but I am having trouble formally proving that it converges. You can also make another numerical algorithm using the inverse of $f_i(\vec{x})$, which is a bit similar to what you have done in 2 dimensions, but only solving $f_i(\vec{x}) = a_i$ for one $i$ at a time causes the other $f_j(\vec{x}) \neq a_j$ so the algorithm needs a few iterations before it converges. I'm glad you think this is a hard problem as I have spent a month trying to solve it! :-) – questiondude Sep 05 '21 at 10:19
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    "strictly increasing" is not well-defined when the domain doesn't have an order relation – Masacroso Sep 05 '21 at 10:32
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    Each $f_i$ is a function with domain $\Bbb R^n$. Strictly increasing means " $x < y \implies f_i(x) < f_i(y)$", but in this case $x<y$ makes no sense since $x,y \in \Bbb R^n$ – jjagmath Sep 05 '21 at 10:35
  • @amsmath I have moved the specific problem to another question here. Perhaps you could move your comment with a solution for 2 dimensions there? I'm sorry for the trouble but I realized it should be two different questions instead. – questiondude Sep 05 '21 at 15:03

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Monotonicity of functions $f:{\mathbb R}^n \to {\mathbb R}^n$ often refers to the order on ${\mathbb R}^n$ defined by the cone $K=[0,\infty)^n$ via $$ x \le y :\iff y-x \in K \quad (\iff x_k\le y_k, k=1,\dots,n). $$ This is an order relation on ${\mathbb R}^n$, but in case $n\ge 2$ its not a total ordering (for example $(1,0)$ and $(0,1)$ cannot be compared). As usual $$ x<y :\iff x \le y \wedge x\not= y, $$ that is $x < y$ means $x_k\le y_k$ for $k=1,\dots,n$ and there is at least one coordinate $l$, say, with $x_l < y_l$. Moreover sometimes $$ x \ll y :\iff x_k<y_k, k=1,\dots,n $$ is used. For example $f: {\mathbb R}^2 \to {\mathbb R}^2$, $f(x_1,x_2) =(x_1+x_2,x_1+x_2)$ ist strongly increasing in the sense that $$ x < y ~ \Rightarrow ~ f(x) \ll f(y), $$ but $f$ is neither surjective (for example $(0,1)$ cannot be reached by $f$) nor injective ($f(0,1)=f(1,0)$). Note that $f(x) \ll f(y)$ implies $f(x) < f(y)$, so $f$ is strictly increasing.

Gerd
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  • Thanks very much for the elaborate answer! Unfortunately I don't fully understand your notation. What is meant by the symbols "$: \iff$"? And what is meant by $\iff$ inside the parentheses in your first equation? I have edited my original question with what I mean by "strictly increasing". Is that essentially what you are writing here as well? Your counter-example is a really good idea, but please explain why it is not surjective? Would you mind elaborating a bit more in your answer, so I can understand it properly as I'm just an amateur mathematician. Thanks! – questiondude Sep 05 '21 at 14:40
  • $:\iff$ just means "defined as". $y-x \in K$ means $y_k-x_k \ge 0$ for each $k=1,\dots,n$. Its just an equivalent description of the order. $f$ is not surjetive, since both coordinates are the same. For example the vector $(0,1)$ cannot be reached by $f$. Strictly increasing with respect to $<$ means: If $x_j \le y_j$ for all $j$ and $x_k<y_k$ in at least one coordinate then $f_j(x) \le f_j(y)$ for all $j$ and $f_l(x)<f_l(y)$ in at least one coordinate. – Gerd Sep 05 '21 at 14:54
  • Thanks! Please incorporate this into your main answer and I'll mark it as answered. I especially like your simple counter-example, so perhaps you could move that on top of your answer for those of us who are not so good at abstract math? – questiondude Sep 05 '21 at 15:07