A question about Euler characteristic in a space product

Question

I have a question about Euler characteristic.

Thing is that the Euler characteristic of the circumference $S^1$ es zero, that is $\mathcal{X}(S^1) = 0$, because we can consider $E = 1$ (all the circumference), $F = 0, V = 1$ (some chosen point).

Since a close interval $[a,b]$ is homeomorphic to $S^1$ we get that Euler characteristic of $[0,1]$, for example, is also $0$.

And we know that $\mathcal{X}(M \times N) = \mathcal{X}(M)\mathcal{X}(N)$. So for the unity square $C = [0,1] \times [0,1]$ it would be $\mathcal{X}(C) = \mathcal{X}([0,1])\mathcal{X}([0,1]) = 0$.

But we know that $C$ has 4 vertices, 4 edges and 1 face: so $\mathcal{X}(C) = 4 + 1 - 4 = 1$. Then, what is wrong here?

Thanks!

Answer 1

In order that this question actually gets an answer, the thing that is wrong - in line with Mike and Federico's comments - is that $[0,1]$ is not homeomorphic to $S^1$.

Answer 2

Actually a variant of your calculation for the euler characteristic of $[0,1]^2$ shows that they are not homeomorphic!

Indeed, if $X$ is homeomorphic to $Y$, then they have the same euler characteristic.

Check by hand, that $\chi([0,1])=1$ by noting that there are two vertices ($0$-cells if you like) and $1$ edge so the Euler characteristic is $2 \cdot (-1)^0+(-1)^1=2-1=1$.

On the other hand, $\chi(S^1)=0$ since it has one vertex and one edge. This shows that they are not homeomorphic.

You can check by hand (as you did) that $\chi([0,1]^2)=1$ or by using the fact that $\chi([0,1]^2)=\chi([0,1] \cdot \chi ([0,1])=1 \cdot 1$.

The latter argument is nice since it implies that $\chi([0,1]^n)=1$ for any hypercube.