Bourbaki

Lemma 4

Tag 01W0, permanent. Cite it as [Bourbaki, Tag 01W0].

Topologie algébrique, chapter III, § 1. Homotopies, homéotopies

Let X and Y be topological spaces, let φ:X→R+\varphi : X\rightarrow \mathbf{R}_+ be a continuous mapping, and put A=φ−1(0)A =\overset{-1}{\varphi}(0). Let σ:X×I→Y\sigma : X\times \mathbf{I}\rightarrow Y be a homotopy which is fixed on A. The mapping σ′:X×I→Y\sigma ': X\times \mathbf{I}\rightarrow Y which maps (x,s)(x, s) to σ(x,s/φ(x))\sigma (x, s/\varphi (x)) if s<φ(x)s < \varphi (x) and to σ(x,1)\sigma (x,1) if s⩾φ(x)s\geqslant \varphi (x) is continuous.

The mapping σ′\sigma ' is continuous at every point of the closed subset φ−1([1,+∞[)×I\overset{-1}{\varphi}([1,+\infty [)\times \mathbf{I}; it is therefore enough to prove that its restriction to φ−1(I)×I\overset{-1}{\varphi}(\mathbf{I})\times \mathbf{I} is continuous. We may therefore suppose that φ(X)⊂I\varphi (X)\subset \mathbf{I}. Let C and C′C' be the subspaces of X×IX\times \mathbf{I} consisting of pairs (x,s)(x, s) such that s⩽φ(x)s\leqslant \varphi (x) and s⩾φ(x)s\geqslant \varphi (x) respectively. They are closed and cover X×IX\times \mathbf{I}. The mapping σ′\sigma ' is continuous on C′C'; let us prove that it is continuous on C.

Let α:X×I→X×I\alpha : X\times \mathbf{I}\rightarrow X\times \mathbf{I} be the continuous mapping defined by α(x,t)=\alpha (x, t) = (x,tφ(x))(x, t\varphi (x)). Its image is equal to C and σ′∘α\sigma '\circ \alpha is the continuous mapping σ\sigma. Let us prove that α\alpha is a proper map. Consider in fact an ultrafilter U\mathfrak{U} on X×IX\times \mathbf{I} and a point (x,t)∈X×I(x, t)\in X\times \mathbf{I} adherent to the ultrafilter basis α(U)\alpha (\mathfrak{U}). Since pr1∘α=_1\circ \alpha = pr1_1, the ultrafilter basis pr1(U)_1(\mathfrak{U}) on X converges to xx. Since I\mathbf{I} is compact, there exists a point s∈Is\in \mathbf{I} such that the ultrafilter basis pr2(U)_2(\mathfrak{U}) converges to ss. Then U\mathfrak{U} converges to (x,s)(x, s). Since α\alpha is continuous, the ultrafilter basis α(U)\alpha (\mathfrak{U}) converges to (x,sφ(x))(x, s\varphi (x)). Since I\mathbf{I} is separated, we have sφ(x)=ts\varphi (x) =t, whence α(x,s)=(x,t)\alpha (x, s) = (x, t), so that α\alpha is proper (TG, I, p. 75, Theorem 1). It then follows from I, p. 18, example 2 and Proposition 9 that σ′∣C\sigma '|C is continuous.

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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Cites

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Machine translation of the English, by gpt-5.4, gpt-5-6-mini, gpt-5-mini, 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.