Bourbaki

Corollary 1

Tag 0B2A, permanent. Cite it as [Bourbaki, Tag 0B2A].

General Topology, chapter I, § 4. Product of topological spaces

Let (Xi)i∈I,(Yi)i∈I(X_i)_{i \in I}, (Y_i)_{i \in I} be two families of topological spaces with the same set of indices. For each i∈Ii \in I, let fif_i be a mapping of XiX_i into YiY_i. In order that the product mapping f:(xi)→(fi(xi))f : (x_i) \to (f_i(x_i)) of

∏i∈IXi into ∏i∈IYi \prod_{i \in I} X_i \text{ into } \prod_{i \in I} Y_i
should be continuous at a point a=(ai)a = (a_i) it is necessary and sufficient that fif_i is continuous at aia_i for each i∈Ii \in I.

f can be written as x→(fi(pri(x)))x \to (f_i(\mathrm{pr}_i(x))), so that by Proposition 1 the condition is sufficient. Conversely, for each x∈Ix \in I let gxg_x be the mapping of XxX_x into ∏i∈IXi\prod_{i \in I} X_i such that prx(gx(xx))=xx\mathrm{pr}_x(g_x(x_x)) = x_x and pri(gx(xx))=ai\mathrm{pr}_i(g_x(x_x)) = a_i whenever i≠xi \neq x; then gxg_x is continuous at the point axa_x, by Proposition 1. Since fx=prx∘f∘gxf_x = \mathrm{pr}_x \circ f \circ g_x it follows that if ff is continuous at aa, then fxf_x is continuous at axa_x.

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

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Transcribed from the printed volume. Not translated.