The 2-genus surface is not contractible.

Question

I'm trying to prove that the 2-genus surface is not contractible. My first idea was to prove that the circle and the 2-genus are homotopy equivalent using an homotopy retract, but since I don't know how to triangulate the 2-genus surface I'm not able to do this. Then, I have tried to compute the fundamental group of the 2-genus surface, but I haven't studied yet The Seifert-Van Kampen Theorem.

Can someone give me a hint please? I have started studying algebraic topology this year, so my knowledge is small. Thank you.

Answer 1

Hint: are there any non-contractible loops in your surface? By definition, a space is contractible if it is homotopy equivalent to a point. What do homotopy equivalences do to the fundamental group?

Answer 2

Follow the hint of @iwriteonbananas.

First, define a retraction (not a deformation retraction, which would be impossible. Just a retraction) of the surface $\mathbb{S}$ to $S^1$. You can do this by considering the surface embedded in $\mathbb{R}^3$, taking a loop $L$ on the plane $xy$ around the two holes (which will be $S^1$), a point $P$ inside a hole and defining the composition $p\circ \pi$, where $\pi$ is the projection $\mathbb{R}^3 \to \mathbb{R}^2$ and $p$ is the projection to the loop with respect to the point $P$.

Having constructed a retraction, use the fact that $\pi_1( \cdot)$ is functorial to conclude that $i_*$, the map induced by the inclusion $L \hookrightarrow \mathbb{S}$, is injective. This will show that $\pi_1(\mathbb{S},*)$ cannot be trivial.

Answer 3

An "elementary" solution: once you figure out how to triangulate the orientable surface of genus $g$, you can calculate its Euler characteristic $$\chi (\Sigma_g) = 2 - 2g.$$ The Euler characteristic of the point is $1$. This argument relies on the invariance of $\chi$ though.

An annoying remark: the hint that you accepted as an answer is not quite a proof: the fact that there's some non-contractible loop is just as intuitively clear as the fact that the surface is not contractible. A true proof would calculate $\pi_1$, or (co)homology, or Euler characteristics (which is a disguised (co)homological invariant).