Euler characteristic of a singular fiber I am trying to understand Kodaira's classification of fibers. In the table at page 41 of Miranda's book  http://www.math.colostate.edu/~miranda/BTES-Miranda.pdf there is given the Euler number of the fiber (e). 
Can someone tell me how to calculate the Euler number of a fiber? 
I guess I cannot apply the formula $v-e+f$ since there is no face on the fiber. 
For example, I need to find $2$ for $I_2$ and $3$ for $I_3$. 
If I apply the formula to them and take the areas between the lines as faces I get the numbers correctly. However, is this how to calculate it?
Any help will be appreciated.
 A: You can use the formula you mention, as long as you apply it correctly. Let's see how.
One consequence of that formula is that if $X \rightarrow Y$ is a map glueing together two points of $X$, then $e(Y)=e(X)-1$, since $Y$ has a triangulation that is the same as that for $X$ except that two vertices in $X$ have been identified in $Y$.
So now let $F$ be a fibre of type $I_n$ in Kodaira's notation. Remember that means $F$ consists of $n$ copies of $\mathbf P^1$ glued together in a chain. $\mathbf P^1$ is topologically $S^2$, so it has Euler number 2. So if we start with $F_0 = \mathbf P^1 \sqcup \cdots \sqcup \mathbf P^1$ we have Euler number $2+ \cdots +2 = 2n$: starting with one of the $\mathbf P^1$, we then glue the other components on, each to the previous one. We glue on a new copy of $\mathbf P^1$ to the existing curve a total of $n-1$ times; finally we glue the "free end" of the last $\mathbf P^1$ to the first curve in the chain. That's a total of $n$ glueings, so we end up with
$$e(F) = e(F_0)-n = 2n-n=n.$$
