In Floer Homology we want to prove that the Moduli spaces $\mathcal{M}(x^{-},x^{+})$ are finite dimensional manifolds. This is done by expressing them as the zero set of a Fredholm map. First one proves that the set $\mathcal{P}^{1,p}(x^{-},x^{+})$ of maps satisfying exponential decay conditions and the boundary conditions $\lim_{s\rightarrow \pm \infty}u(s,\cdot)=x^{\pm}$ is a Banach manifold . And then we have that $\mathcal{M}({x^{-},x^{+}})$ is the zero set of the map $\mathcal{F}^{H,J}_u(Y):=\Phi_u(Y)^{-1}\mathcal{F}(\exp_u Y)$ where $Y$ is a vector field along $u$ and $\mathcal{F}$ is the Floer equation.

I have been wondering for a while why we use this map $\mathcal{F}^{H,J}_u$. We will have that it's linearization will be a nice Fredholm operator since we can use Unitary trivializations of the bundles $u^*(TM)$ to show that $d(\mathcal{F}_{u}^{H,J})_0$ is conjugated to an operator of the form

$$D_u: W^{1,p}(\mathbb{R}\times S^1,\mathbb{R}^{2n})\rightarrow L^p(\mathbb{R}\times S^1,\mathbb{R}^{2n})$$ $$Y\mapsto \partial_sY+J_0\partial_t Y+S(s,t)Y$$

and these operators are very well understood and in particular we can compute it's Fredholm index. Is there any other reason other than this to why we use this map $F^{H,J}_u$?

I am asking this because if we want extra conditions on our Moduli spaces of maps $\mathcal{M}(x^{-},x^{+})$, which we call this space $\mathcal{M}^k$, such that we can prove that $\mathcal{P}^{1,p,k}$ is a banach manifold then I am not sure we can use the map $\mathcal{F}^{H,J}_u$ to prove that $\mathcal{M}^k$ is a banach manifold since this might not be the zero section of $\mathcal{F}^{H,J}_u$ due to the fact that $\exp_u Y$ might not need to verify the new boundary condition. For example the boundary condition could be a Lagrangian boundary conditions an in Lagrangian Floer Homology.

I am not sure if in this case we need to pick a metric $\exp$ so that all of this works or if there is another thing going on .

Any insight is appreciated, thanks in advance.