Consider a set of functions $\mathcal{F}$ on $E$ where $E \subset\mathbb{R}^k$ - e.g. the class of $L_1$ functions on $[0,1]$ - and endow it with a suitable metric $d$ that makes it Polish.

Consider a partition $\{\mathcal{F}_i, i\in I\}$ (say at most countable) of $\mathcal{F}$ and denote by $d_i$ the restriction of $d$ to $\mathcal{F}_i$. Is it true that the topological space $(\mathcal{F},\tau_d)$ - where $\tau_d$ is the $d$-metric topology - coincides with the disjoint union topological space $\coprod_{i \in I} (\mathcal{F}_i,\tau_{d_i})$ - where $\tau_{d_i}$ is the $d_i$-metric topology on $F_i$ - only if $\mathcal{F}_i \in \tau_d$, for all $i \in I$?

If yes, can we also conclude that the Borel $\sigma$-algebras induced by open sets under the two topological structures coincide only if $\mathcal{F}_i \in \tau_d$, for all $i \in I$? Or would it be sufficient to have $\mathcal{F}_i \in \sigma_B(\tau_d)$, for all $i \in I$, where $\sigma_B(\tau_d)$ is the Borel $\sigma$-algebra induced by $\tau_d$?

I guess that there's probably a link with an old question:

Is there a "disjoint union" sigma algebra?

My questions rise from asking myself whether one could construct a $\sigma_B(\tau_d)$ \ $\sigma_B(G)$-measurable map from $\mathcal{F}$ to some metric space $G$, with Borel $\sigma$-algebra $\sigma_B(G)$, by starting from some continuous maps $\phi_i:\mathcal{F}_i \mapsto G$, $i \in I$, and then appeal to the universal property of coproducts.