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I'm late, but I think the answer can be given in a simpler fashion.

A principal bundle is in particular a fibre bundle, but there exists no residual choice of what the fibre should be — it must be $G$ by definition. To me, the most direct way of recovering the choice of a fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle (with fibres $F$) like $P\times F$ and then dividing out the unwanted old fibres $G$: $$E:=(P\times F)/G\quad\text{meaning we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$

I'm late, but I think the answer can be given in a simpler fashion.

A principal bundle is in particular a fibre bundle, with the extra restriction that the fibre has to be a group $G$. To me, the most direct way of again relaxing to an arbitrary fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle (with fibres $F$) like $P\times F$ and then dividing out the unwanted groupy fibres $G$: $$E:=(P\times F)/G\quad\text{meaning we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$

I'm late, but I think the answer can be given in a simpler fashion.

A principle bundle is in particular a fibre bundle, but there exists no residual choice of what the fibre should be - it must be $G$ by definition. To me, the most direct way of recovering the choice of a fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle (with fibres $F$) like $P\times F$ and then dividing out the unwanted old fibres $G$: $$E:=(P\times F)/G\quad\text{meaning we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$

I'm late, but I think the answer can be given in a simpler fashion.

A principal bundle is in particular a fibre bundle, but there exists no residual choice of what the fibre should be — it must be $G$ by definition. To me, the most direct way of recovering the choice of a fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle (with fibres $F$) like $P\times F$ and then dividing out the unwanted old fibres $G$: $$E:=(P\times F)/G\quad\text{meaning we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$

I'm late, but I think the answer can be given in a simpler fashion.

A principle bundle is in particular a fibre bundle with no choice of what the fibre should be - it must be $G$ by definition. To me, the most direct way of recovering the choice of a fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle with fibres $F$ like $P\times F$ and then dividing out the unwanted fibres $G$: $$E:=(P\times F)/G\quad\text{i.e. we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$

I'm late, but I think the answer can be given in a simpler fashion.

A principle bundle is in particular a fibre bundle, but there exists no residual choice of what the fibre should be - it must be $G$ by definition. To me, the most direct way of recovering the choice of a fibre $F$ in $P(M,G)$ is to first enlarge $P$ to be a trivial fibre bundle (with fibres $F$) like $P\times F$ and then dividing out the unwanted old fibres $G$: $$E:=(P\times F)/G\quad\text{meaning we identify}\quad (u,f)\sim(ug,g^{-1}.f).$$ This is like if you want $5$ to be $7$, you boldly write $$7=(5\times 7)/5.\qquad\text{(sorry)}$$