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.


1. 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 $i \in I$?
2. 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?](https://mathoverflow.net/questions/54923/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. 



Topologies and Borel $\sigma$-fields on disjoint unions
*Remarks and hints, not a solution*  

**Question 1** A disjoint union $\mathcal F = \bigcup\_{i \in I} \mathcal F\_i$ in a metric space has the disjoint union topology if and only if all sets $\mathcal F\_i$ are open in $\mathcal F$. Is that question 1? The answer is yes. Why not try to prove it?  

In particular, whether $\mathcal F$ is a space of functions, or whether the metric is Polish, or indeed whether $\mathcal F$ it is metrizable at all: these do not come into it.


**Question 2** No, it could happen that some $\mathcal F\_i$ is not open, but the two Borel sigma-algabras coincide anyway. Perhaps you can find an example where there are just two sets $\mathcal F\_i$.


