Lemma 2

If $(f_i)_{i \in I}$ is a family of continuous mappings $f_i : X_i \to Y_i$, each of which satisfies condition d) of Theorem 1, then the product mapping

also satisfies d).

Let $U$ be an ultrafilter on $X = \prod_i X_i$, and let $y = (y_i)$ be a point of $Y = \prod_i Y_i$ such that $f(U)$ converges to $y$. This means that each of the ultrafilter bases $\operatorname{pr}_i(f(U)) = f_i(\operatorname{pr}_i(U))$ converges to $y_i$ (§ 7, no. 6, Proposition 10, Corollary 1). By virtue of condition d), for each $i \in I$ there exists $x_i \in X_i$ such that $f_i(x_i) = y_i$ and $\operatorname{pr}_i(U)$ converges to $x_i$; but then $U$ converges to $x = (x_i)$ (loc. cit.) and we have $f(x) = y$. This completes the proof of Lemma 2 and hence of Theorem 1.