Proposition 4

Tag 05CV, permanent. Cite it as [Bourbaki, Tag 05CV].

*With the above notation, there exists one and only one isomorphism u of $\mathscr{E}_{\tau}$ onto $\mathscr{E}'$ such that $u \circ s = s'$ .

(f, g) = (f, e) \cdot_{\tau} (e, g) = i(f) \cdot_{\tau} s(g).

Therefore, if

u

is a solution to the problem, of necessity

u(f, g) = i'(f).s'(g)

, whence the uniqueness of

u

. We prove the existence. We write

u(f, g) = i'(f).s'(g)

. Then

\begin{align*} u(f, g).u(f', g') &= i'(f)s'(g)i'(f')s'(g') \\ &= i'(f)(s'(g)i'(f')s'(g)^{-1})s'(g)s'(g') \\ &= i'(f)i'(\tau(g, f')).s'(g)s'(g') \\ &= i'(f.\tau(g, f')).s'(gg') \\ &= u((f, g) \cdot_{\tau} (f', g')). \end{align*}

Therefore, $u$ is a homomorphism of $F \times_{\tau} G$ into $E'$ . Obviously $u \circ i = i'$ , $p' \circ u = p$ and $u \circ s = s'$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

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