Euler characteristic of a surface

Question

It is known that a closed orientable surface of genus $g$ has Euler characteristic $2-2g$. According to this, the open disc being of genus $0$ should have Euler characteristic $2$, but this contradicts the fact that the disc is contractible so has a Euler characteristic $1$. Thank you for your clarification!!

Answer 1

First of all we need to clarify what is your definition of the Euler characteristic of a space $X$. For me, the Euler characteristic of a finite dimensional CW complex can be defined to be the alternating sum over the number of cells in each dimension. As it turns out, for a finite dimensional CW complex $X$ the Euler characteristic is also equal to

$$\sum_{i=0}^k (-1)^i \textrm{rank} H_i(X)$$

For the closed orientable surface $M_g$ of genus $g$, we have $H_0(X) = \Bbb{Z}$, $H_1(X) = \Bbb{Z}^{2g}$, $H_2(X) = \Bbb{Z}$ and so

$$\chi(M_g) = 1 - 2g + 1 = 2- 2g.$$

Now for the open/closed disk $X$ (it does not matter which one) we see that $H_0(X) = \Bbb{Z}$, and all higher homology groups vanish because it is contractible. Hence

$$\chi(X) = 1.$$

Answer 2

It is known that a closed orientable surface of genus $g$ has Euler characteristic $2-2g$. Being closed is essential here. Indeed, consider for example the open disc. This surface is orientable of genus zero but is not compact hence it is not a closed surface so it does not verify the formula $\chi=2-2g$ and in fact the actual $\chi$ of the open disc is $1$ since the disc is contractible. Moreover the closed disc is compact but it has a boundary so it is not a closed surface neither and hence the formula does not apply; in fact the closed disc is also contractible and has $\chi=1$.