Bourbaki

Example 2

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

Topologie algébrique, chapter III, § 2. Homotopie et chemins

Soit nn an integer ⩾0\geqslant 0. Let PnP_n denote the set of monic polynomials P∈P\in C[X]\mathbf{C}[X] of degree nn, endowed with the topology for which the mapping (c0,...,cn−1)↦Xn+cn−1Xn−1+⋯+c0(c_0, . . . , c_{n-1})\mapsto X^n+c_{n-1}X^{n-1}+\cdots +c_0 is a homeomorphism of Cn\mathbf{C}^n onto PnP_n. The mapping pp of Cn\mathbf{C}^n into PnP_n defined by p(z1,...,zn)=(X−z1)...(X−zn)p(z_1, . . . , z_n) = (X-z_1). . .(X-z_n) is continuous. The symmetric group Sn\mathfrak{S}_n operates on Cn\mathbf{C}^n by permutation of the factors and pp defines by passing to the quotient a homeomorphism of Cn/Sn\mathbf{C}^n/\mathfrak{S}_n onto PnP_n (TG, VIII, p. 22, prop. 1, I, p. 23, cor. 1 and TG, VIII, p. 20). We therefore deduce the following statement:

Let I be an interval of R\mathbf{R}, let (c0,...,cn−1)(c_0, . . . , c_{n-1}) be a sequence of continuous mappings from I into C\mathbf{C}, let aa be a point of I and let (z1,...,zn)(z_1, . . . , z_n) be a sequence of complex numbers such that one has (X−z1)...(X−zn)=Xn+(X-z_1). . .(X-z_n) = X^n+ cn−1(a)Xn−1+⋯+c0(a)c_{n-1}(a)X^{n-1}+\cdots +c_0(a). There exists a sequence (λ1,...,λn)(\lambda_1, . . . , \lambda_n) of continuous mappings from I into C\mathbf{C} such that one has λi(a)=zi\lambda_i(a) =z_i for 1⩽i⩽n1\leqslant i\leqslant n and (X−λ1(t))...(X−λn(t))=Xn+cn−1(t)Xn−1+⋯+c0(t)(X-\lambda_1(t)). . .(X-\lambda_n(t)) = X^n+c_{n-1}(t)X^{n-1}+\cdots +c_0(t) for every t∈It\in I.

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-5-6-mini, gpt-5-mini, gpt-5-6, gpt-5.4, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.