Question on continuous function between squares

Question

Let $Q\subset \mathbb{R}^2$ be the square determined by the points $(0,0)$, $(4,0)$, $(4,4)$, $(0,4)$ and let $f:Q\rightarrow Q$ be any continuous function which fixes the four points $(0,0)$, $(4,0)$, $(4,4)$, $(0,4)$.

Denote by $d\subset Q$ the diagonal between $(0,0)$ and $(4,4)$ and by $f(d)$ its image by $f$: it is clear that $f(d)$ is a continuos path in $Q$ between $(0,0)$ and $(4,4)$. The path $f(d)$ splits $Q$ in many parts, denote by $Q'_0$ the one containing $f(4,0)$.

Question: is it true that $f(2,1)$ must be contained in $Q'_0$?

My guess would be "yes", I'll explain my reasoning.

The function $f:Q\setminus d\rightarrow Q\setminus f(d)$ is still continuous and thus must send connected sets to connected sets. Since $f$ fixes $(4,0)$ then the lower triangle must be sent entirely in $Q_0'$.

Is it right?

Answer 1

well, no. The square can be folded... For some point $ 0 < x+y \leq 4,$ define $$ T = x+y $$ and then $$ \lambda = \frac{y}{T} = \frac{y}{x+y}. $$ The fraction $\lambda = 0$ refers to the point $(T,0). $ The fraction $\lambda = 1$ refers to the point $(0,T). $ Then let $$ f(x,y) = (T,0) + \left( 16 \lambda^3 - 24 \lambda^2 + 9 \lambda \right) (-T,T). $$ With some effort, this can be written out as two rational functions of $x,y.$

I have not written it down yet, the behavior for $4 < x+y < 8$ is a mirror image of what I said above. FRIDAY: For some point $ 4 \leq x+y \leq 8,$ define $$ T = x+y $$ but then $$ \lambda = \frac{y+4-T}{8-T}. $$ The fraction $\lambda = 0$ refers to the point $(4,T-4). $ The fraction $\lambda = 1$ refers to the point $(T-4,4). $ Then let $$ f(x,y) = (4,T-4) + \left( 16 \lambda^3 - 24 \lambda^2 + 9 \lambda \right) (T-8, 8-T). $$

For any $x = y$ we get $\lambda = 1/2$ and $f(T/2,T/2) = (T/2,T/2).$

For your $(2,1)$ we get $T=3,$ then $\lambda = 1/3,$ then $$ f(2,1) = (3,0) + \left( \frac{25}{27} \right) (-3,3)=\left( \frac{2}{9}, \frac{25}{9} \right) . $$

Answer 2

Your argument holds if $f$ is one-to-one, but if $f(Q \setminus d) \cap (Q \setminus f(d)) \neq \emptyset $, the statement is not true in general. Your proposed proof assumes that $f$ restricts to a function of $Q \setminus d $ to $Q \setminus f(d)$, but if that is not true, we can find a counter-example.

For example, we can design a continuous $f$ from $Q = [0, 4] \times [0, 4]$ to itself with the following properties, where $A = (0, 0)$, $B = (0, 4)$, $C = (4, 4)$ and $D = (0, 4)$ are the vertices of the square $Q$ and where $Z = (3, 0)$ (a point lying on the lower horizontal edge $AB$ to the east of the point $X = (2, 1)$ that we are trying to isolate from $B$ in $f(Q)$):

1. $f$ maps the rectangle $[0, 3] \times [0, 4]$ to the triangle $\triangle AZD$.
2. $f$ maps the rectangle $[3, 4] \times [0, 4]$ to the triangle $\triangle ZBC$.
3. $f$ fixes $A, B, C, D$ and $Z$.

So the image of such an $f$ is contained in the union of the two triangles $\triangle AZD$ and $\triangle ZBC$. Given such an $f$, the image $f(d)$ of the diagonal $d$ contains the point $Z$ so removing $f(d)$ will separate $f(B) = B = (4, 0)$ from $f(X) = f((2, 1)) \in \triangle AZD$.

To construct an example of such an $f$, we can scale each vertical interval $\{x\} \times [0, 4]$ in $Q$ to map it onto the vertical interval through $\triangle AZD$ or $\triangle ZBC$ that it meets:

$$ f((x, y)) = \left\{% \begin{array}{l@{\quad}l} (x, \frac{3-x}{3}y) & \mbox{if $x \in [0, 3]$} \\ (x, (x - 3)y) & \mbox{if $x \in [3, 4]$} \end{array}\right. $$