Why is the Euler characteristic for the circle $0$? The Euler characteristic of an edge is $-1$, so what makes the circle different?
 A: Actually, the Euler characteristic of an edge is $+1$ according to the most common definition. In terms of "triangulations", you can describe the circle as the union of two edges which meet at two vertices so the Euler characteristic is $2 - 2 = 0$. In terms of homology groups, we have $\dim H_0(S^1) = \dim H_1(S^1) = 1$ and $\dim H_i(S^1) = 0$ for $i \geq 2$ so the Euler characteristic is $1 - 1 = 0$.
A: You can think of an (open) edge as one $1$-dimensional cell, while a circle as a disjoint union of one $0$-dimensional cell (that is, a point) and one $1$-dimensional cell.   Since a common definition of the Euler characteristic is $k_0-k_1+k_2-\cdots$, where $k_i$ denotes the number of $i$-dimensional cells in a cell decomposition of the underlying topological space, we get the answers $-1$ and $0$ for the edge and circle respectively. 
A: Thinking about it informally and geometrically, if you cut a circle at $A$, then you get two same vertices $A$,  an edge and no face. So 
$$
\chi = V - E + F = 1 - 1 + 0 = 0.
$$
