# Homeomorphism of a Genus-2 Surface

Does there exist a homeomorphism from a genus-2 surface, the connected sum of 2 tori, to two circles, $S^1$, intersecting at a point? Intuitively it seems that the double torus can be squeezed into the desired shape continuously, but I am not able to prove this. If this mapping makes sense could somebody sketch the idea of why this is so, and in the null case explain what goes wrong?

If it turns out that this homeomorphism exists, can I say more about the mapping? Namely, is the mapping also a diffeomorphism?

Here is a picture of the genus-2 surface under consideration

and here is a picture of two circles touching at a point

• They're not even locally homeomorphic. The genus 2 surface is locally homeomorphic to $\Bbb R^2$, and the "wedge of two circles" (your bottom picture) is locally homeomorphic, except at the intersection point, to $\Bbb R$. In fact, they're not even homotopy equivalent (looking either at homology or the fundamental group will show this).
– user98602
Aug 27, 2014 at 17:37
• Are you asking for a homeomorphism or a deformation retract? Also, is it the solid 2-torus? Aug 27, 2014 at 17:37
• Taking the question at face value, the answer is (as others have said) no: If you remove any point from the $2$-torus the resulting space is connected; if you remove the common point of the two wedged circles, the resulting space has two connected components. Aug 27, 2014 at 17:58
• @QuangHoang it was the deformation retract that I was after. Michael's answer got to the heart of what I was considering. Aug 27, 2014 at 18:12

The two-holed torus is a topological two-manifold (i.e. a surface), so each point has a neighbourhood homeomorphic to an open subset of $\mathbb{R}^2$; this property is often called locally Euclidean of dimension two. If the two circles meeting at a point (often called the wedge sum of two circles, written $S^1\vee S^1$) were homeomorphic to the two-holed torus, then it would also be locally Euclidean of dimension two, but this is not the case (can you see why?). Therefore, the two spaces are not homeomorphic.
I'm guessing that your question arose from the fact that both spaces have two 'holes'. While it is not true that the two spaces are homeomorphic, there is a way of relating the two spaces if you instead consider the solid two-holed torus (i.e. you include the points on the inside). In that case, we can deform the solid two-holed torus to the wedge sum of two circles. More precisely, we say that $S^1\vee S^1$ is a deformation retract of the solid two-holed torus.