Is the circle homeomorphic to a $6$ petal rose?

Question

I am trying to explain that a circle is homeomorphic to a $6$-petal rose with the standard topology of $\mathbb R^2$, for that i'll need to explain (just in words, it is not necessary to come up with a formula) the homeomorphic function that maps one into the other.

So starting from the center of the rose, a series of continous changes can start to push neighboring points of the rose to the border of the circle, in the end every point should be at an equal $r$ distance from the center of the rose. The same thing goes for the inverse, starting with six equidistance points in the circumference, a series of continous chanages should push the their neighboing points following the trajectory of the petals to the center of the flower. The map and the inverse should be continous because an arbitrary open ball of a certain radius in one figure, should be map to another open ball of a certain radius in the other one. The problem is in the center of the flower where the points converge, there is no simil open set in the circle, would that mean that there is no bijection relationship between their topologies?, even though they both share the same. Is the open ball at the center of the flower different from the other ones?, so despite that the and its inverse are continous, it is not homeomorphic?, or the function is not continous?. I am stuck and can't find an explanaition that makes sense, so any insight would be greatly appreciated.

Kind Regards

Answer 1

They are not homeomorphic. If you remove the center of the rose, what remains is disconnected. However, if you remove a point from a circle, what remains stays connected.

Answer 2

Actually it's not true, because if you remove the center of the rose you split the rose in 6 parts (connected components) while if you remove a point from the circle you have just 1 part.

Answer 3

Other brilliant commenters have shown that they are not homeomorphic. I wanted to add to this by showing a continuous function $f: \mathcal{S}^1 \to \Bbb{R}^2$ that maps the circle to a six-petal flower: $$f(\theta) = (|\sin (3\theta)|, \theta)$$ Where $\theta$ specifies a point on the circle in a continuous fashion (for example, in a $\Bbb{R}^2$ setting, it could be the angle in polar coordinates). The function itself outputs the point in polar coordinates.

An interesting interpretation of the above is, by restricting the map on only its output on the plane (thus turning it into a surjection), that this is a quotient map. In particular this map generates a space where six points on the circle belong to the same equivalence class while the others keep their topology intact - that is, a six-petal flower.