Is the action functional "locally well defined up to an additive constant"? I'm looking at an arbitrary symplectic manifold $(P, \omega)$, and the action functional on the set of contractible loops $\Lambda P:$ $$A: \Lambda P \to \mathbb{R}, \hspace{5pt} A(z) = \int_{D^2} \bar{z}^*\omega,$$where $\bar{z}$ is some extension of $z: S^1 \to P$ to $D^2$, which exists by contractibility.
Of course, this function is not well defined, because the value of $A$ will almost certainly depend on the choice of extension $\bar{z}$ (unless for example $\pi_2(P) = 0$). However, according to Floer in his paper "The unregularized gradient flow of the symplectic action," this function is locally well-defined, up to an additive constant, so we can talk about its derivative $dA$.

For a fixed $z \in \Lambda P$, how does one choose the neighborhood $z \in U \subset \Lambda P$ on which $A$ can be well-defined, with a suitable, coherent choice of extensions of loops in $U$?

For what it's worth, here's my attempt, even though I don't think it needs to be that complicated.
Choose $z$ and a "small," contractible neighborhood $U$ of $z$ (small enough so that all loops in $U$ are homotopic to $z$). Choose also an arbitrary extension $\bar{z}$ of $z$ to $D^2$. For any other $z' \in U$, choose a homotopy $h: z \simeq z'$. Since $D^2 \times \{0\} \cup S^1 \times I$ is a retract of $D^2 \times I$, there exists a map $\tilde{h}: D^2 \times I \to P$ with $\tilde{h}(x, 0) = \bar{z}(x)$ and $\tilde{h}|_{S^1 \times I} = h$. Thus I can define a lift of  $\bar{z}' := \tilde{h}|_{D^2 \times \{1\}}$, and set $$A(z') := \int_{D^2} \bar{z}'^* \omega.$$However, this is no better than choosing an arbitrary extension, because the homotopy $h$ was arbitrary. So instead of looking at the extension problem

I should be looking at all of the homotopies simultaneously, something like

where ideally I'd be able to choose the homotopies to depend continuously (even smoothly) on $z' \in U$. Since $U \times D^2 \times I$ is contractible, there's definitely an extension up to homotopy, but that's not what I need. If $U \times D^2 \times I$ had a CW-complex structure, I could find a solution, but it doesn't in general because $U$ lives in an infinite-dimensional manifold, and they don't admit CW-structures, according to https://mathoverflow.net/questions/357477/cw-structure-on-infinite-dimensional-manifolds/357522#357522.
P.S: I know that people usually work with some quotient of the universal cover of $\Lambda P$ instead etc, but I'm just interested to see if this approach can work too.
 A: Answering your edit: This follows from your choice of neighbourhood $U$, namely from its contractibility. Two continuous maps $f,g \in C(X,Y)$ lie in the same path component (using a suitable topology) if and only if they are connected. Since $U$ is contractible, so are $U_{[z]}$ and $U_{[\tilde{z}]}$, and hence $[\bar{z}]$ ($[\tilde{z}]$) and $[\bar{z}']$ ($[\tilde{z'}]$) are homotopic.
A: I think I figured it out: I was wrong to look at extensions $\bar{z}$ themselves rather than the homotopy classes. I will assume that $P$ is path connected, so that $\Lambda P$ is also.
Denote by $\widetilde{\Lambda P}$ the universal covering of $\Lambda P$. This is, as a set, for some choice of $p \in P$, $$\widetilde{\Lambda P} = \{ \overline{z}: I \to \Lambda P \hspace{3pt} | \hspace{3pt} \overline{z}(0) = c_p\}/ \simeq (\textrm{rel } \{0,1\}),$$and this can be reinterpreted as $$\widetilde{\Lambda P} = \{\overline{z}:D^2 \to P \hspace{3pt} | \hspace{3pt} \overline{z}(0) = p\}/\simeq (\textrm{rel } S^1 \cup \{0\}),$$with the covering map $p: \widetilde{\Lambda P} \to \Lambda P$ being nothing but the restriction to $S^1$. Then we can define $$\widetilde{A}: \widetilde{\Lambda P} \to \mathbb{R}, \hspace{5pt} \widetilde{A}[\overline{z}] := \int_{D^2} \overline{z}^* \omega.$$If $\overline{z}, \overline{z}'$ belong to the same equivalence class, then, if $h: \overline{z} \simeq \overline{z}'$, we have $$\int_{D^2} \overline{z}^* \omega - \int_{D^2} \overline{z}'^* \omega = \int_{\partial(D^2 \times I)} (h^* \omega)|_{\partial(D^2 \times I)} = \int_{D^2 \times I} d h^* \omega = 0,$$so $\widetilde{A}$ is well-defined.
Now, take $z \in \Lambda P$, and a neighborhood $z \in U \subset \Lambda P$ such that $$ p^{-1}(U) = \bigsqcup_{[\overline{z}] \in p^{-1}(z)} U_{[\overline{z}]},$$ and $p: U_{[\overline{z}]} \to U$ is a smooth diffeomorphism. Choosing some such $[\overline{z}]$ (this is the additive constant), we can define $$A: U_z \to \mathbb{R}, \hspace{5pt}, A(z') := \widetilde{A}[\overline{z}'],$$where $[\overline{z}']$ is the unique element of $U_{[\overline{z}]} \cap p^{-1}(z').$
EDIT: Actually, I still don't know how to prove the "up to an additive constant" part. Suppose I have two homotopy classes of extensions of $z$, $[\overline{z}]$ and $[\widetilde{z}]$. If I take $z' \in U_z$ and $[\overline{z'}]$ and $[\widetilde{z'}]$, two extensions of $z'$ in $U_{[\overline{z}]}$ and $U_{[\widetilde{z}]}$, respectively, then I need to show that $$\int_{D^2} \overline{z'}^* \omega - \int_{D^2} \widetilde{z'}^* \omega = \int_{D^2} \overline{z}^* \omega - \int_{D^2} \widetilde{z}^* \omega.$$
