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If there is a continuous function from the closed unit disk to itself such that it is identity map on boundary, must it admit a fixed point in the interior of the disk?

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  • $\begingroup$ You might want to give a link to the other question and perhaps the 'in the interior' part... $\endgroup$
    – copper.hat
    Mar 2, 2015 at 7:23

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No. Consider the set of vertical line segments which are intersections of lines $x=c$ with the closed unit disc $x^2+y^2 \le 1.$ Each of these segments goes from a lower circle boundary point $P_c$ to an upper one $Q_c.$ Along each such segment we can choose a mapping which moves all the interior points but keeps the endpoints fixed. An example of such a map for the interval $[0,1]$ is $h(t)=t^2.$ It would seem not too difficult to, in this way, choose the collection of maps on the intervals $P_c Q_c$ in such a way that the resulting pieced together mapping is continuous, and it would fix the points on the boundary of the unit disc by construction. Also by construction there would not be interior fixed points inside the unit disc.

An explicit map: First note that for $a\le b$ one can parametize the closed interval $[a,b]$ as $(1-t)a+tb$ where $t \in [0,1].$ Here $t=0$ goes with the smaller endpoint $a$ while $t=1$ goes with endpoint $b$.

We can apply this to the unit disc $x^2+y^2 \le 1$ by letting its points be of the form $(x,y(t))$ where $$y(t)=(1-t)[-\sqrt{1-x^2}] + t [\sqrt{1-x^2}].\tag{1}$$ Note that for $x= \pm 1$ this correctly gives $y=0$ for any $t,$ so this is a continuous parametrization (in the new variables $x$ and $t$) of the disc and its interior. the points where $t=0$ correspond to the lower boundary of the disc, while points with $t=1$ go with the upper boundary.

Now to define the map in these terms, we map the point $(x,y(t))$ onto the point $(x,y(t^2)).$ In general the value of $t$ uniquely determines the point $(1-t)a+tb$ in the interior of the closed interval $[a,b]$, and also at the endpoints provided $a<b$. This assures us that our mapping has no fixed points interior to the disc, while it is clearly continuous. It fixes every point on the boundary of the disc, since $0^2=0$ and $1^2=1$ at the two extremes of $[0,1]$.

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  • $\begingroup$ Under this construction, is it continuous? cuz the boundary is fixed. $\endgroup$
    – David
    Mar 2, 2015 at 8:23
  • $\begingroup$ @David I have abandoned the previous stereographic construction since I think it failed to fix the boundary points of the unit circle. Please have a look at this adjusted attempt, which does need more detail. $\endgroup$
    – coffeemath
    Mar 2, 2015 at 8:44
  • $\begingroup$ @David I had some time, so I filled in some details giving hopefully an understandable explicit map. I still feel there must be a "simpler" such map, but can't think of an explicit one so far... $\endgroup$
    – coffeemath
    Mar 2, 2015 at 13:12

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