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Let $C_1, C_2 \subset \mathbb{R}^2$ be concentric circles in the plane. Suppose that $C_1$ bounds $C_2$. Let $f: C_1 \rightarrow C_2$ be a map such that for some $y \in C_2$, $f(x) = y$ for all $x \in C_1$ and that $x$ and $y$ must be connected by a path that is entirely contained in the interior of the region bounded by $C_1$ and that does not intersect $C_2$. Can $f$ be continuous?

EDIT:

I should probably phrase this differently. If $f$ is defined as above, can the sequence of paths under the condition above be continuous?

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  • $\begingroup$ Yes, like concentric circles $C_i$ centered on x-axis. $ (x-h)^2+y^2 = r^2$, for arbitrary $h,r$. $\endgroup$ – Narasimham Dec 5 '15 at 6:55
  • $\begingroup$ I don't think it really matters where they're centered. Just as long as $C_2$ is contained in $C_1$. $\endgroup$ – user296029 Dec 5 '15 at 6:57
  • $\begingroup$ You write "$f(x)=y$ for all $x\in C_1$"...so $f$ is a constant function? $\endgroup$ – Eric Wofsey Dec 5 '15 at 6:57
  • $\begingroup$ In any case, I don't understand the condition you're trying to impose by saying that $x$ and $y$ can be connected by a certain path...there exists such a path for every $x\in C_1$ and every $y\in C_2$. $\endgroup$ – Eric Wofsey Dec 5 '15 at 6:59
  • $\begingroup$ Yes, every point on $C_1$ maps to the same point on $C_2$ by $f$. But the restriction that $x$ and $y$ be connected by a path in the interior of the region bounded by $C_1$ that does not intersect $C_2$ is why I think that this must violate continuity. $\endgroup$ – user296029 Dec 5 '15 at 6:59
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Every constant map is continuous, so in particular your map is continuous.

However, you cannot choose such paths from each $x\in C_1$ to $y$ such that they give a continuous map $F:C_1\times [0,1]\to B$, where $B$ is the closed annulus between $C_1$ and $C_2$. To see this, let $p:B\to C_1$ be the radial projection and consider the composition $pF:C_1\times [0,1]$. Then $pF(x,0)=x$ for all $x$ and $pF(x,1)=p(y)$ for all $x$, so $p$ is a homotopy from the identity map $C_1\to C_1$ to a constant map. Since $C_1$ is not contractible, no such homotopy exists.

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  • $\begingroup$ I guess more specifically I should say that the sequence of paths generated can't be continuous. Or rather does there exist a continuous constant function $f$ that satisfies my condition. $\endgroup$ – user296029 Dec 5 '15 at 7:03
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Yes, for instance the radial map $$ x^2 + y^2 = h\, y $$ where $ h <1 $ maps all circles on x-axis to its dilated counterparts.

EDIT1:

Or if non-intersection among circles is very strict, the $\tau $ circles in bipolar co-ordinate map or grid (Fig 181):

$$ x^2- 2 x h + y^2 +T^2 =0 $$

The circles are distinct, non touching, same constant tangent length $L$ , variable set constant $h$ for each circle... but this example is from non-euclidean hyperbolic geometry, out of the given tag.

Bipolar_Tau_Circles

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