In the (wonderful) book by C. Birkenhake and H. Lange *Complex Abelian Varieties* we can find the following result, see Corollary 4.3.4 page 77. It is stated in any dimension $g \geq 2$, but let us consider only the case of abelian surfaces for the sake of simplicity.

>**Proposition.** Let $f \colon X \to Y$ be an isogeny of abelian surfaces and let $D$ be a positive definite and irreducible divisor on $Y$. Then $f^*D$ is also irreducible.

The proof starts as follows: 

*Assume the contrary, then $f^*D$ is a sum of effective divisors $D_1+ \cdots + D_n$. But necessarily $D_i \cdot D_j=0$ and $D_i$ is numerically equivalent to $D_j$ for all $i \neq j$, the map $f$ being an étale Galois covering $\ldots$* 

and then a contradiction is reached by using the Nakai-Moishezon theorem.

Now, I do not understand the part $D_i \cdot D_j=0$ for all $i \neq j$. 

If $D$ is smooth then this is immediate: in fact, an étale cover of a smooth curve must be smooth, in particular its irreducible components do not intersect. 

However, if $D$ is singular (and in the Proposition there is no smoothness assumption) the same argument do not work. Indeed, it is well-known that there are irreducible, $1$-nodal curves with a connected, étale double covering consisting of two copies of the normalization, see for instance example 10.6 in Chapter III of Hartshorne's *Algebraic Geometry*. The two components intersect at two points, that are the two nodes of the covering.

So my question is

>**Q.** Is the result stated in the Proposition above also true when $D$ is singular? If not, what is a counterexample? Or maybe am I missing something?