# Circle and heart homeomorphic?

Is a circle and heart homeomorphic to one another?

Intuitively, I can picture that the one can be "morphed" into the other by bending and stretching and not breaking. But I am unsure if that is correct?

This is not an assignment or anything, I am just thinking about it in general.

Can anyone please confirm or reject my reasoning above and show it to me (algebraically or through a sketch or anything) in order to give me a nice explanation?

This is the picture that got me thinking about it • At least the usual heart shape, e.g., upload.wikimedia.org/wikipedia/commons/thumb/f/f1/… is, yes. We'd usually refer to a "filled" circle as a disk, though. May 8, 2015 at 16:39
• @Travis - can you please provide me with the correct terminology and reasoning as an answer, so I can accept and give you the credit? :). May 8, 2015 at 16:41
• Yes, they are. One way to prove this would be to write down an explicit equation for a map between them and show that it and its inverse are continuous. May 8, 2015 at 16:41
• @Dillon I'm off to bed, but if I see no one has answered this by tomorrow, I'll at least write up a sketch. Often writing down explicit homeomorphisms can be unpleasant, even for two spaces that are "obviously" homeomorphic. Let me recommend for you the problem of showing that a disk and ("filled") square are homeomorphic, which captures some of the key issues in this problem, but which is probably a little easier to handle. May 8, 2015 at 16:45
• @Dillon I believe convex is not a topological property... See here: en.wikipedia.org/wiki/Topological_property On pp67 you get a good idea of homeomorphisms: books.google.at/… However, proving that something is not homeomorph topological properties are usually the way to go :) May 8, 2015 at 18:50

Here's a possible approach. Let $$D=\{(x,y)\in\mathbb R^2\mid x^2+y^2\leq1\}$$ be the closed unit disk.

For any pair of functions $f,g:[-1,1]\to\mathbb R$ such that $f(-1)=g(-1)$, $f(1)=g(1)$ and $f(t)<g(t)$ for all $t\in(-1,1)$, let $$A(f,g)=\{(x,y)\in\mathbb R^2\mid f(x)\leq y\leq g(x)\}$$ be "the area between their graphs". (Most candidates for "the heart shape" can be described as $A(f,g)$ for a suitable choice of $f,g$.) Now, I claim that any such set $A(f,g)$ is homeomorphic to $D$. We shall prove this in several steps.

Lemma 1. $A(f,g)$ is homeomorphic to $A(0,g-f)$. (Here $0$ is the zero function, defined by $0(x)=x$ and $g-f$ is defined by $(g-f)(x)=g(x)-f(x)$.)

Proof. The homeomorphism $h:A(f,g)\to A(0,g-f)$ is simply $$h(x,y)=(x,y-f(x))$$ which obviously well-defined and is continuous, because $f$ is. Its inverse is given by $$h^{-1}(x,y)=(x,y+f(x))$$ and we're done. $\square$

Lemma 2. Suppose $k_i:[-1,1]\to\mathbb R$, $i=1,2$ are any two functions such that $k_i(-1)=k_i(1)=0$ and $k_i(t)>0$ for $t\in(-1,1)$. Then $A(0,k_1)$ is homeomorphic to $A(0,k_2)$.

Proof. Again, we may define an explicit homeomorphism $h:A(0,k_1)\to A(0,k_2)$, this time by the formula $$h(x,y)=\left(x,\frac{k_2(x)}{k_1(x)}y\right)$$ for $x\in(-1,1)$ and $h(-1,0)=h(1,0)=0$. This is continuous for $x\in(-1,1)$, since $k_1$ and $k_2$ are continuous and products and quotients of continuous functions are continuous (the latter wherever the denominator is nonzero). But $h$ also continuous at the points $(\pm1,0)$, since $\frac{y}{k_1(x)}\in[0,1]$ for all $(x,y)\in A(0,k_1)$, while $k_2(x)$ goes to $0$ as $x$ approaches $\pm1$. So the two limits $\lim_{(x,y)\to(\pm1,0)}h(x,y)$ exist and equal the corresponding function values. By the same argument, the inverse $$h^{-1}(x,y)=\left(x,\frac{k_1(x)}{k_2(x)}y\right)$$ is continuous. So $h$ is indeed a homeomorphism. $\square$

Proposition. $A(f,g)$ is homeomorphic to $D$.

Proof. By Lemma 1, $A(f,g)$ is homeomorphic to $A(0,g-f)$. By Lemma 2, $A(0,g-f)$ is homeomorphic to $A(0,2k)$ with $k(x)=\sqrt{1-x^2}$. By Lemma 1 again, $A(0,2k)$ is homeomorphic to $A(-k,k)=D$. Therefore $A(f,g)$ is homeomorphic to $D$. $\square$

If you want an explicit homeomorphism, simply calculate the composition of all the maps used. (By the way, the proofs are even simpler if you work with an open unit disk and "open heart" instead: this way you don't have to analyse the points $(\pm1,0)$ separately.)

To obtain (polygonal version of) a heart, you could use e.g. $$f(x)=|x|-1$$ and $$g(x) = \frac12-\left||x|-\frac12\right|,$$ but I'm sure you can come up with a better ("round" version of a) heart yourself and the same argument will work.

In any case, it is probably helpful to spend some time trying to visualize what each of the maps does.