I have a function $f:R^n_+\rightarrow R^n$ for which I want to show the following:

$$\exists c\in R^n_+ \quad \forall i,j\,\,f_i(c)=f_j(c)$$ Where $f_i (c)$ are the different coordinates of $f$.

$f$ has the following properties:

- $\frac{\partial f_i}{\partial c_j}>0 \Leftrightarrow i=j$, furthermore the partial derivatives are never $0$.
- $\lim_{c_i\rightarrow \infty} f_i(c) = \infty$
- $\lim_{c_j\rightarrow \infty} f_i(c) = A_i$, $A_i$ here does not depend on $j$ (and of course $j\neq i$).
- $\forall t\in R_+ \quad f(c) = f(t\cdot c)$

For $n=2$ this is very easy, basically just the intermediate value theorem. For higher dimension it gets more complicated. The idea is the following: If there does not exist such a $c$, then the image of $f$ is contained in $R^n-\{x\in R^n|x_1 = x_2\cdots=x_n\}$, which is topicologically different from $R_+^n$ (the preimage). All we have to show is something like we have a circle around this line, that can't then be contracted.

A bit more formally: We define $\tilde{e}_i = [1,\dots,1,1/\epsilon,1,\dots,1]$, where $1/\epsilon$ is at the position $i$. With these points we have $f(\tilde{e}_i)\approx [A_1, A_2, \cdots, B_i,\dots,A_n]$, with $B_i$ being a huge number. We can then define path $p_{ij}:[0,1]\rightarrow R_+^n$, $p_{ij}(t) = t\cdot \tilde{e}_i +(1-t)\cdot \tilde{e}_i$. Then we connect all the path $f(p_{ij})$. These path will then from a closed path "around" the removed line $\{x\in R^n|x_1 = x_2\cdots=x_n\}$. This path could not be contracted if this line were not in the image of $f$. Therefore we have at least one such point.

Questions:

- Does simply connected suffice also for higher dimensions or do I need algebraic topology?
- Is there a way to proof that this point is unique?
- Is there a more beautiful way to proof this? In its current version it's quite a mess.

algebraictopology? Is your $f$ algebraic? 2. Do you have an example of such an $f$? 3. Quoting from en.wikipedia.org/wiki/Simply_connected_space : "The image of a simply connected set under a continuous [or even entire analytic -- I.P.] function need not be simply connected. Take for example the complex plane under the exponential map: the image is C - {0}, which is not simply connected". $\endgroup$ – Iosif Pinelis Jun 1 '18 at 14:119more comments