Maslov index equal to $2$ implies that the disk is not multiply covered

Question

In the paper Floer cohomology of lagrangian intersections and pseudo-holomorphic disks in page $1009$ the authors claim that a map $w:D^2\rightarrow S^2$ with $w|_{\partial D^2}\subset L$, where $L$ is the equator, and such that $\mu(w)=2$, where $\mu$ is the maslov index of the map, cannot be multiply covered and hence it will be injective in the interior of $D^2$.

I have tried proving this , but I am getting nowhere.

Following the paper "Relative frames on $J$-holomorphic curves" the definition of multicovered that I found is that there exists a simple disk $v:(D,\partial D)\rightarrow (S^2,L) $ and a surjective map $p:(D,\partial D)\rightarrow (D,\partial D)$ continuous on $D$, holomorphic on the interior , satisfying $p^{-1}(\partial D)=\partial D$ and $u=v\circ p$ thus $[u]=m[v]$ in relative homology with $m\geq 1$.

And so using this and the fact that the minimal maslov number is $2$ will force $m=1$ and $[u]=[v]$.

However from here I am not sure out to prove that $u$ is bijective in the interior of $D^2$.

Any help or reference where I can look this up is appreciated, thanks in advance.

Answer 1

First, I assume you want $w$ to be holomorphic (or else it isn't true).

If the target is really $S^2$ with boundary on the equator $L$ then this should be easy to prove directly. Take a point $p$ not on $L$. We can define an intersection number between $w$ and $p$ using the intersection pairing between $H_2(S^2,L)$ and $H_0(S^2\setminus L)$. By positivity of intersections, any point $q$ in the preimage of $p$ under $w$ contributes positively to this intersection number (with multiplicity 1 if and only if $w$ is transverse to $p$ at $q$). Let's write $[N]$ and $[S]$ for the Northern and Southern hemispheres considered as (Maslov 2) classes (in fact generators) in $H_2(S^2,L)$. The class of $w$ is $a[N]+b[S]$ for some $a$ and $b$ with $a+b=1$ because $\mu(w)=2$. The intersection number of $w$ with $p$ is $a$ if $p\in N$ and $b$ if $p\in S$. Since both of these must be nonnegative and sum to 1, we get $[w]=[N]$ or $[S]$. Moreover, the preimage of each point in $N$ (or $S$) is a single point and the derivative of $w$ is nonzero there to get intersection multiplicity 1 at $p$.

Moreover, the preimage of any point in $N$ (or $S$) has the same intersection number with the disc, so $w$ is really just a bijective parametrisation of a hemisphere.

In fact, this works for any embedded circle on the sphere (the argument didn't depend on areas). I'm sure there are other, simpler, ways to see it that don't explicitly invoke relative homology; this is just one that sprang to mind.