Open set in the manifold

Question

we have $(X,F,\mu)$ be the measure space and let $M_{\mu}=\{p:X\to R \text{ measurable }: p>0 \mu -a.e and \int p d\mu =1\}$. For each positive density $p\in M_{\mu}$ we define the Orlicz space $L^{\Phi}(p.\mu)$. Then we have $B_{p}=\{u\in L^{\Phi_1}(p.\mu): E_{p}[u]=0\}$ where $E_{p}$ is the expectation. Now we have for each $p\in M_{\mu}$ let $V_{p}$ be the open unit ball in $B_{p}$ that is $V_{p}=\{u \in B_{p}: ||u||_{\Phi_1,p}<1\}$. we define a map $$e_{p}:V_{p}\to M_{\mu}$$ such that $$e_{p}(u)= e^{u-log E_{p}[e^u]}p.$$ This map is one-one. The range of $e_p$ is $U_{p}$. The inverse image of $e_{p}$ on $U_{p}$ is the function $$s_{p}:q\in U_{p}\mapsto log\left(\frac{q}{p}\right)-E_{p}\left[log\left(\frac{q}{p}\right)\right]$$ Now it is given that $(U_p,s_p)$ for $p\in M_{\mu}$ is a chart. So basically here we are giving the manifold structure to $M_{\mu}$ now my question is that when we say chart that means $U_{p}$ will be the open set of $M_{\mu}$ but how do we show that? I mean $M_{\mu}$ is just a set there is no topology that we have defined on it then how we will show it. Please someone help me in this. Thanks.