Bourbaki

Example 1

Tag 01S2, permanent. Cite it as [Bourbaki, Tag 01S2].

Topologie algébrique, chapter I, § 6. Espaces simplement connexes

Let X be a simply connected topological space and let ff be a continuous function from X into C∗\mathbf{C}^*. Recall (I, p. 101, example 6) that the mapping z↦ezz\mapsto e^z makes C\mathbf{C} a covering of C∗\mathbf{C}^*. By prop. 3, there exists a continuous function h:X→Ch: X\rightarrow \mathbf{C} such that f(x)=eh(x)f(x) =e^{h(x)} for every x∈Xx\in X. If h′:X→Ch': X\rightarrow \mathbf{C} is another continuous function such that f(x)=eh′(x)f(x) =e^{h'(x)} for every x∈Xx\in X, there exists an integer q∈Zq\in \mathbf{Z} such that h′=h+2πiqh'=h+ 2\pi iq.

Analogously, let nn be an integer >0>0; the mapping z↦znz\mapsto z^n makes C∗\mathbf{C}^* into a covering of itself (I, p. 101, example 5). There therefore exists a continuous function kk from X into C∗\mathbf{C}^* such that k(x)n=f(x)k(x)^n=f(x) for every x∈Xx\in X, for example the function k(x)=eh(x)/nk(x) =e^{h(x)/n}. If k′:X→C∗k': X\rightarrow \mathbf{C}^* is another continuous function such that k′(x)n=f(x)k'(x)^n=f(x) for every x∈Xx\in X, there exists an nen^e root of unity μ∈C\mu\in \mathbf{C} such that k′=μkk'=\mu k.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Machine translation of the English, by gpt-5.4, against glossary version 34. Not checked by a person.