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.
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 $$e(F) = e(F_0)-n = 2n-n=n.$$