Compactly Choquet-complete spaces III: products and continuous almost open images

The compactly Choquet-complete spaces were introduced in [1], and we had started to study them last time and the month before. We had shown that compactly Choquet-complete spaces are completely Baire, and that the class of compactly Choquet-complete spaces contains the LCS-complete spaces (the G_δ subspaces of locally compact sober spaces), and were closed under taking closed subspaces.

We continue our list, and we will show that the class of compactly Choquet-complete spaces is closed under countable products and under images by continuous almost open surjective maps.

The case of countable products will be interesting in that we will need to simulate countably many games in parallel; and if infinitely many games must be simulated, then we will start them later and later, so that only finitely many games are being simulated at each individual step of the game.

As a reminder, the strong Choquet game is played between two players, α and β. Player β starts the game by picking a point x₀ and an open neighborhood V₀ of x₀. Then α replies with an open neighborhood U₀ of x₀ included in V₀; β picks a possibly different point x₁ in U₀, an open neighborhood V₁ of x₁ included in U₀; α replies with an open neighborhood U₁ of x₁ included in V₁; β picks a possibly different point x₂ in U₁, an open neighborhood V₂ of x₂ included in U₁, and so on. After both players have played for as many turns as there are natural numbers, α is declared to win if ∩_n ∈ N U_n (which is also equal to ∩_n ∈ N V_n) is non-empty. Otherwise, β wins. We say that α wins compactly if and only if ∩_n ∈ N U_n is a non-empty compact set Q_∞ and the sets U_n form a base of open neighborhoods of Q; α wins convergently if and only if it wins compactly and Q_∞ is the upward closure of a point x_∞.

Each player plays according to a strategy: σ for α, τ for β. Player α finds U_n as σ(x₀, V₀, U₀, x₁, V₁, U₁, ···, x_n, V_n), and player β finds the pair x_n, V_n as τ(x₀, V₀, U₀, x₁, V₁, U₁, ···, x_n–1, V_n–1, U_n–1). A strategy σ for α that wins compactly (resp., convergently) against every possible strategy τ that β uses is compactly winning (resp., convergently winning). A space is compactly Choquet-complete (resp., convergence Choquet-complete) if and only if α has a compactly (resp., convergently) winning strategy.

A stationary strategy for α is one that only depends on the last pair x_n, V_n played by β, and we write σ(x_n, V_n) instead of σ(x₀, V₀, U₀, x₁, V₁, U₁, ···, x_n, V_n) in that case. A space is stationary compactly Choquet-complete (resp., stationary convergence Choquet-complete) if and only if α has a stationary compactly (resp., stationary convergently) winning strategy.

Products of compactly Choquet-complete spaces

We will see that countable products Π_i ∈ I X_i of compactly Choquet-complete spaces X_i are compactly Choquet-complete. Finite products are pretty easy to deal with: we play compactly winning strategies in each space X_i in parallel. In the infinite countable case, we need a new trick.

Let us see right away why uncountable products do not preserve compact Choquet-completeness. We will write points of a product Π_i ∈ I X_i in bold face, as x, and open subsets as U, V, W, similarly. The product topology on Π_i ∈ I X_i is generated by open rectangles, which are products Π_i ∈ I U_i where each U_i is open in X_i and U_i=X_i for all i except for finitely many. We will call support of such an open rectangle the (finite) collection of indices i ∈ I such that U_i≠X_i.

Counterexample A. Let I be an uncountable index set, and let X_i be the space of real numbers R, with its usual metric topology. Taking R is not important; the important thing is that the spaces X_i are not compact (and non-empty). Let us imagine that α has a compactly winning strategy σ on Π_i ∈ I X_i = R^I. We define a strategy τ for β as follows: initially, β produces x₀, V₀ where x₀ is an arbitrary point of Π_i ∈ I X_i, and V₀ ≝ Π_i ∈ I X_i. Given a history x₀, V₀, U₀, x₁, V₁, U₁, ···, x_n–1, V_n–1, U_n–1, where n≥1, β produces x_n ≝ x_n–1, and an open rectangle V_n containing x_n and included in U_n–1 (we don’t care which). Since σ is compactly winning, Q_∞ ≝ ∩_n ∈ N U_n is compact and non-empty. Since U_n ⊆ V_n ⊆ U_n–1 for every n≥1, Q_∞ is also equal to ∩_n ∈ N V_n. Now each V_n is an open rectangle; let J_n be its support. Since V_n+1 ⊆ U_n ⊆ V_n for every n ∈ N, J_n+1 ⊇ J_n. Let J be the union ∪_n ∈ N J_n. Every J_n is finite, so J is countable, and therefore I–J is uncountable, in particular non-empty. By definition, each V_n is a product of open subsets of each X_i, and all those at indices in I–J (hence in I–J_n) are equal to X_i. Therefore we can write V_n as V‘_n × Π_{i ∈ I–J} X_i where V‘_n is an open rectangle in Π_i ∈ J X_i. In particular, Q_∞ = ∩_n ∈ N V‘_n × Π_{i ∈ I–J} X_i. We fix an index i in I–J, and we write π_i for projection of Π_i ∈ I X_i onto X_i. This is a continuous map, so π_i[Q_∞] is compact. But π_i[Q_∞] = π_i[∩_n ∈ N V‘_n × Π_{i ∈ I–J} X_i] = X_i (the latter holds because ∩_n ∈ N V‘_n × Π_{i ∈ I–J} X_i = Q_∞ is non-empty), which is not compact.

This is to be contrasted with the fact that arbitrary products of Choquet-complete spaces are Choquet-complete, a result I will leave as an exercise.

In order to show that finite or countable products of compactly Choquet-complete spaces are compactly Choquet-complete spaces, we will need an extension of Wallace’s theorem, already mentioned in the November 2023 post (Appendix A): given two topological spaces X and Y, a compact subset Q of X, a compact subset Q’ of Y, and an open neighborhood W of Q × Q’, there is an open neighborhood U of Q in X and an open neighborhood V of Q’ in Y such that U × V ⊆ W. The extension to come works for arbitrary products, not just finite or countable products.

Lemma B.  Let (X_i)_i ∈ I be a family of topological spaces, and let Q_i be a compact subset of X_i, one for each i ∈ I.  For every open neighborhood W of Π_i ∈ I Q_i in Π_i ∈ I X_i, there is a family (U_i)_i ∈ I of open subsets of X_i, one for each i ∈ I, such that Q_i  ⊆ U_i for every i ∈ I, U_i=X_i for all i except for finitely many, and Π_i ∈ I U_i  ⊆ W.

Proof. If some Q_i is empty, then this is obvious: we take the empty set for U_i, X_j for U_j for every j≠i, and this fits the bill.  Hence we will assume that every Q_i is non-empty.  This will only be important near the end of the proof.

We first prove this when I is finite.  Let X₁, …, X_n be n topological spaces, Q_i be a compact subsets of X_i, one for each i ∈ {1, ···, n}, and W be an open neighborhood of Q₁ × ··· × Q_n in X₁ × ··· × X_n.  We show the claim by induction on n.  If n=0, then Q₁ × ··· × Q_n and X₁ × ··· × X_n contain exactly one element, and the claim is trivial.  Otherwise, Q₁ × ··· × Q_n–1 is compact in X₁ × ··· × X_n–1 by Tychonoff’s theorem (Theorem 4.5.12 in the book), and (Q₁ × ··· × Q_n–1) × Q_n is included in W.  By Wallace’s theorem, there is an open subset U of X₁ × ··· × X_n–1 and an open subset U_n of X_n such that Q₁ × ··· × Q_n–1 ⊆ U, Q_n ⊆ U_n, and U × U_n ⊆  W.  By induction hypothesis, there are open subsets U₁, …, U_n–1 of X₁, …, X_n–1 such thatQ₁ ⊆ U₁, …, Q_n–1 ⊆ U_n–1 and U₁ × ··· × U_n–1 ⊆ U.  In particular, U₁ × ··· × U_n ⊆ U × U_n ⊆  W.

Second, we prove the lemma in the general case, where I is no longer assumed to be finite. For short, let us call open rectangle any product Π_i ∈ I U_i  of open subsets U_i of X_i, one for each i ∈ I, such that U_i=X_i for all i except for finitely many. We will say that such an open rectangle is supported on a given set J of indices included in I if and only if the indices i such that U_i≠X_i are all in J.

By definition of the product topology, W is a union of open rectangles.  Π_i ∈ I Q_i is compact by Tychonoff’s theorem (Theorem 4.5.12 in the book) and is included in W, hence is included in finitely may open rectangles R₁, …, R_n included in W.  Each of these rectangles is supported on a finite subset of I, and by taking their union, we obtain a finite subset J of I such that R₁, …, R_n are all supported on J.  Let π be the projection map from Π_i ∈ I X_i onto Π_j ∈ J X_j.  This is a continuous map, so the image π[Π_i ∈ I Q_i] is compact in Π_j ∈ J X_j; alternatively, π[Π_i ∈ I Q_i] is the product Π_j ∈ J Q_j (I will let you check this; this is where we need every Q_j to be non-empty!). The image π[R₁ ∪ ··· ∪ R_n] is equal to π[R₁] ∪ ··· ∪ π[R_n], which is a union of open rectangles in Π_j ∈ J X_j.  Hence π[R₁ ∪ ··· ∪ R_n] is open in Π_j ∈ J X_j, and it contains the product of compact sets π[Π_i ∈ I Q_i] = Π_j ∈ J Q_j.  By the first part of this proof (the case of finite products), there are open subsets U_j of X_j, one for each j ∈ J, such that Q_j  ⊆ U_j for every j ∈ J, and Π_j ∈ J U_j ⊆ π[R₁ ∪ ··· ∪ R_n].  We let U_i≝X_i for all i in I–J, and this completes the proof.  Then Q_i  ⊆ U_i for every i ∈ I, U_i=X_i for all i except for finitely many, and Π_i ∈ I U_i = π^–1 (Π_j ∈ J U_j) ⊆ π^–1 (π[R₁] ∪ ··· ∪ π[R_n]) = π^–1 (π[R₁]) ∪ ··· ∪ π^–1(π[R_n]), and the latter is equal to R₁ ∪ ··· ∪ R_n (because each rectangle among R₁, …, R_n is supported on J), which is included in W. ☐

We use this to show that finite products compactly Choquet-complete spaces are compactly Choquet-complete. The case of countable products will require an additional trick, and we prefer to show a simpler version first.

Proposition C. The class of compactly Choquet-complete spaces is closed under finite products.  Similarly for stationary compactly Choquet-complete spaces, for convergence Choquet-complete spaces, for stationary convergence Choquet-complete spaces.

Proof. Let (X_i)_i ∈ I be a finite family of topological spaces, and for each i ∈ I, let σ_i be a compactly winning strategy for player α on X_i (resp., and stationary).

We build a (stationary if every σ_i is stationary) compactly winning strategy for α on Π_i ∈ I X_i as follows.  For every point x of Π_i ∈ I X_i, we will write x[i] for its component i, namely its image under projection π_i, in X_i; there will be sequences of points x_n below, and they should not ne confused with the component x[n].  For every open subset W of Π_i ∈ I X_i, we will also write W[i] for the projection of W onto component i, namely {x[i] | x ∈ W}.  For every open subset W of Π_i ∈ I X_i, W[i] is open.  We stress the fact that, in general, Π_i ∈ I W[i] ≠ W, except (and this is an important exception) when W is an open rectangle.

Given that β has just played x_n, V_n, α finds an open rectangle W_n containing x_n and included in V_n.  For each i∈ I, W_n[i] is an open subset of X_i, and Π_i ∈ I W_n[i] = W_n… because W_n is an open rectangle.  Player α then computes U_n ≝ Π_i ∈ I σ_i(x₀[i], V₀[i], U₀[i], x₁[i], V₁[i], U₁[i], ···, x_n[i], V_n[i]) (in the stationary case, just Π_i ∈ I  σ_i(x_n[i], V_n[i])).  In other words, player α simulates playing in each space X_i in parallel. This product U_n is open because the index set I is finite. And U_n is the value of the desired strategy σ on the history x₀, V₀, U₀, x₁, V₁, U₁, ···, x_n, V_n (resp, on x_n, V_n in the stationary case). For future reference, we note that U_n = Π_i ∈ I U_n[i].

For every i ∈ I, since σ_i is compactly winning, Qⁱ_∞ ≝ ∩_n ∈ N U_n[i] is compact and non-empty in X_i, and the sets U_n[i] form a base of open neighborhoods of Qⁱ_∞ when n varies over N.  Let Q_∞ ≝ Π_i ∈ I Qⁱ_∞.  By Tychonoff’s theorem (Theorem 4.5.12 in the book), Q_∞ is compact in Π_i ∈ I X_i.  It is also non-empty.  For every n ∈ N, we have Qⁱ_∞ ⊆ U_n[i] for every i ∈ I, so Q_∞ ⊆ U_n.

It remains to show that for every open neighborhood W of Q_∞ in Π_i ∈ I X_i, there is an n ∈ N such that U_n ⊆ W.  By Lemma B (our extension of Wallace’s theorem), there is an open rectangle U (hence U = Π_i ∈ I U[i]) such that Qⁱ_∞ ⊆ U[i] for every i ∈ I, and U ⊆ W.  For every i ∈ I, Qⁱ_∞ ⊆ U[i]; since the sets U_n[i] form a base of open neighborhoods of Qⁱ_∞ when n varies over N, there is a natural number n_i such that U_ni[i] ⊆ U[i].  We use the fact that I is finite in order to find a natural number n larger than or equal to n_i for every i ∈ I.  Then U_n[i] ⊆ U[i] for every i ∈ I. Hence U_n= Π_i ∈ I U_n[i] ⊆ Π_i ∈ I U[i] = U ⊆ W.

When all the strategies σ_i are stationary, the strategy we have built is stationary.  Hence products of stationary compactly Choquet-complete spaces are stationary compactly Choquet-complete.

If the spaces X_i are (stationary) convergence Choquet-complete, then every Qⁱ_∞ is of the form ↑xⁱ_∞ for some point xⁱ_∞ of X_i, so Q_∞ = ↑x where x ≝ (xⁱ_∞)_i ∈ I.  This shows that Π_i ∈ I X_i is (stationary) convergence Choquet-complete.  ☐

The case of infinite countable products is more difficult.

We cannot let α play in all spaces X_i at once in parallel (using strategy σ_i), because the products of open sets U_n[i] would in general fail to be open. Fortunately, the open rectangle W_n that player α built in the proof of Proposition C above… is an open rectangle, hence has a finite support J, and we only need to apply the strategies σ_i in the finitely many spaces X_i such that i ∈ J. For the other indices i, we let U_n[i] be X_i, and then U_n = Π_i ∈ I U_n[i] will be an open rectangle.

This solves one problem. Unfortunately, for any given index i, this strategy may decide to never play inside space X_i, if i happens to belong the support of no open rectangle W_n played during the game. In that case, U_n[i] will be equal to X_i for every n ∈ N, and therefore Qⁱ_∞ ≝ ∩_n ∈ N U_n[i] will be equal to X_i, which may fail to be compact. The solution is to force α to eventually play in every X_i. We will maintain a finite set J_n of indices i ∈ I in which we will let α play. J_n will increase (or stay the same) when n increases, and we will force it contain n if n is not already in it.

Theorem D.  The class of compactly Choquet-complete spaces is closed under countable products.  Similarly for convergence Choquet-complete spaces.

Proof.  Since the case of finite products has already been dealt with in Proposition C, we assume a countably infinite (X_i)_i ∈ N of topological spaces, indexed by N, and for each i ∈ N, we let σ_i be a compactly winning strategy for player α on X_i (resp., and stationary).

We build a memoryful compactly winning strategy for α on Π_i ∈ N X_i as follows. The memory will consist of a set J_n–1 of positions, as described above (starting with J_–1 ≝ ∅), and also, for each j ∈ J_n–1, of the time (= step number in the game) n_j at which we started playing in space X_j.

At step n of the game, where β has just played x_n, V_n, player α finds an open rectangle W_n containing x_n and included in V_n, as in the proof of Proposition C.  Let J_n be the union of J_n–1 (from the memory), the support of W_n, and {n} (in order to force α to play in space X_n). We memorize J_n for the next step. For every j ∈ J_n–J_n–1, we also memorize n_j ≝ n: we will start playing in those spaces X_j right now, at step n. Then α plays U_n ≝ Π_i ∈ I U_n[i] where:

• U_n[i] ≝ X_i if i ∈ N–J_n (we are not playing in that X_i yet);
• U_n[j] ≝ σ_j(x_nj[j], V_nj[j], U_nj[j], x_nj+1[j], V_nj+1[j], U_nj+1[j], ···, x_n[j], V_n[j]) for every j ∈ J_n: not σ_j(x₀[j], V₀[j], U₀[j], x₁[j], V₁[j], U₁[j], ···, x_n[j], V_n[j])! We need to simulate a game that starts at step n_j, hence σ_j only sees the points and open sets played at step n_j and later, but cannot see any previous item in the history. (It does not matter that β can see previous parts of the history.)

Since J_n is finite, U_n ≝ Π_i ∈ N U_n[i] is an open rectangle. For every i ∈ N, the time n_i at which α starts playing in space X_i exists. Indeed, n ∈ J_n, so we even have n_i≤n. Since σ_i is compactly winning, Qⁱ_∞ ≝ ∩_n≥ni U_n[i] is compact and non-empty in X_i, and the sets U_n[i] form a base of open neighborhoods of Qⁱ_∞ when n varies over the natural numbers ≥ n_i.  Let Q_∞ ≝ Π_i ∈ N Qⁱ_∞.  By Tychonoff’s theorem (Theorem 4.5.12 in the book), Q_∞ is compact in Π_i ∈ N X_i.  It is also non-empty.  For every n ∈ N, for every i ∈ N, we have Qⁱ_∞ ⊆ U_n[i] if n≥n_i (since Qⁱ_∞ ≝ ∩_n≥ni U_n[i]) but also if n<n_i (since then U_n[i]=X_i), so Q_∞ ⊆ U_n.

It remains to show that for every open neighborhood W of Q_∞ in Π_i ∈ N X_i, there is an n ∈ N such that U_n ⊆ W.  By Lemma B, there is an open rectangle U (hence U = Π_i ∈ N U[i]) such that Qⁱ_∞ ⊆ U[i] for every i ∈ I, and U ⊆ W.  Since U is an open rectangle, the collection of indices i ∈ N such that U[i]≠X_i is finite, call it J.  It is easy to see that J_n contains {0, 1, …, n} for every n ∈ N. Hence J ⊆ J_n for every natural number larger than or equal to m, where m is any fixed natural number larger than or equal to every element of J. For every j ∈ J, Q^j_∞ ⊆ U[j]; since the sets U_n[j] with n≥n_j form a base of open neighborhoods of Q^j_∞, there is a natural number m_j≥n_j such that U_mj[j] ⊆ U[j].  We use the fact that J is finite in order to find a natural number n larger than or equal to m_j for every j ∈ J, and larger than or equal to m.  Then U_n[j] ⊆ U[j] for every j ∈ J. For every i ∈ N–J, U_n[i] ⊆ U[i] holds trivially because U[i]=X_i, by definition of J. Hence U_n= Π_i ∈ N U_n[i] ⊆ Π_i ∈ I U[i] = U ⊆ W.

If the spaces X_i are convergence Choquet-complete, then every Qⁱ_∞ is of the form ↑xⁱ_∞ for some point xⁱ_∞ of X_i, so Q_∞ = ↑x where x ≝ (xⁱ_∞)_i ∈ I.  This shows that Π_i ∈ I X_i is convergence Choquet-complete.  ☐

Images under continuous (almost) open maps

The images of Choquet-complete spaces under continuous open surjective maps are Choquet-complete (Exercise 7.6.19 in the book), and similarly with convergence Choquet-complete spaces (Exercise 7.6.20 in the book).  We adapt the technique to compactly Choquet-complete spaces, and we also take the opportunity to show that this still holds for continuous almost open maps.  A function f : X → Y is almost open if and only if ↑f[U] is open for every open subset U of X; it would be open if and only if f[U] itself, without the need for taking upward closures, were open for every open subset U of X.

Theorem E. The following classes are closed under images by continuous almost open surjective maps: Choquet-complete spaces, convergence Choquet-complete spaces, compactly Choquet-complete spaces and their stationary variants.

Proof. Let f : X → Y be a continuous almost open surjective map.  Since f is surjective, using the Axiom of Choice, there is a function s : Y → X such that f o s = id_Y.  The function s is not in general continuous.

We start with stationary strategies.  Given a stationary strategy σ for α on X, we build a stationary strategy σ’ for α on Y by letting σ'(y_n, V_n) be U_n ≝ ↑f[U’_n] where U’_n ≝ σ(s(y_n), f^–1(V_n)). We have y_n ∈ U_n = ↑f[U’_n] because s(y_n) ∈ U’_n, so y_n = f(s(y_n)) ∈ f[U’_n]; and U_n = ↑f[U’_n] ⊆ V_n, too: since U’_n ⊆ f^–1(V_n), we obtain f[U’_n] ⊆ V_n hence also ↑f[U’_n] ⊆ V_n since V_n is upwards-closed.

If σ is winning (in the sense used for plain Choquet-completeness), then ∩_n ∈ N U’_n is  non-empty.  We pick a point x_∞ in ∩_n ∈ N U’_n, and then f(x_∞) is in every f[U’_n], hence in every U_n = ↑f[U’_n], hence in ∩_n ∈ N U_n.  If σ is convergent winning, then we can pick x_∞ so that the sets U’_n form a base of open neighborhoods of x_∞.  For every open neighborhood U of f(x_∞), x_∞ is in f^–1(U), and therefore U’_n is included in f^–1(U) for some n ∈ N.  Then f[U’_n] is included in U, and since U is upwards-closed, U_n = ↑f[U’_n] is included in U.

If σ is compactly winning, then ∩_n ∈ N U’_n is a compact non-empty set Q’_∞, and the sets U’_n form a base of open neighborhoods of Q’_∞.  Let Q_∞ ≝ ∩_n ∈ N U_n.  We claim that Q_∞ = ↑f[Q’_∞].  That is not completely obvious, since direct images under f do not in general commute with unions, and similarly for upward closures. Since Q’_∞ ⊆ U’_n for every n ∈ N, ↑f[Q’_∞] ⊆ ↑f[U’_n] = U_n for every n ∈ N, so ↑f[Q’_∞] ⊆ Q_∞. Conversely, since Q_∞ is an intersection of upwards-closed sets, Q_∞ is itself upwards-closed, i.e., saturated.  In order to show that Q_∞ ⊆ ↑f[Q’_∞], it suffices to show that every open neighborhood U of ↑f[Q’_∞] contains Q_∞.  Since ↑f[Q’_∞] ⊆ U, Q’_∞ is included in f^–1(U).  Since the sets U’_n form a base of open neighborhoods of Q’_∞, one of them is included in f^–1(U).  Then f[U’_n] ⊆ U, hence U_n = ↑f[U’_n] ⊆ U since U is upwards-closed.  Now Q_∞ ⊆ U_n, so Q_∞ ⊆ U.  This concludes our argument showing that Q_∞ ⊆ ↑f[Q’_∞], and therefore that Q_∞ = ↑f[Q’_∞].  In passing, we have shown that for every open neighborhood U of ↑f[Q’_∞], there is an n ∈ N such that U_n = ↑f[U’_n] ⊆ U.  Therefore the sets U_n form a base of open neighborhoods of Q_∞.  It follows that σ’ is compactly winning.

In the general case, we are given a strategy σ for α on X, not necessarily stationary.  We build a memoryful strategy σ’ for α on Y as follows.  The memory will consist of a list of open subsets U’₀, U’₁, ···, U’_n–1 of X.  Given a history y₀, V₀, U₀, y₁, V₁, U₁, ···, y_n, V_n on Y, we form a corresponding history s(y₀), f^–1(V₀), U’₀, s(y₁), f^–1(V₁), U’₁, ···, s(y_n), f^–1(V_n) on X.  Notice that we completely ignore U₀, U₁, ···, U_n–1, and we use the memory U’₀, U’₁, ···, U’_n–1  instead.  We apply σ to that history, obtaining an open subset U’_n of X that contains s(y_n) and is included in f^–1(V_n).  Then we add U’_n to the memory, this obtaining U’₀, U’₁, ···, U’_n–1, U’_n as new memory, and we return U_n ≝ ↑f[U’_n] as the value of σ'(y₀, V₀, U₀, y₁, V₁, U₁, ···, y_n, V_n) we were seeking to define.

Now let us assume that β plays according to some strategy τ on Y.  We obtain a mirror strategy τ’ for β on X by letting τ'(x₀, V’₀, U’₀, x₁, V’₁, U’₁, ···, x_n–1, V’_n–1, U’_n–1) be  x_n, V’_n where y_n, V_n ≝ τ(f(x₀), ↑f[V’₀], ↑f[U’₀], f(x₁), ↑f[V’₁], ↑f[U’₁], ···, f(x_n–1), ↑f[V’_n–1], ↑f[U’_n–1]), x_n ≝ s(y_n) and V’_n ≝ f^–1(V_n).

The key point is that: (*) for every open subset V of Y, ↑f[f^–1(V)] = V.  Indeed, in one direction for every y ∈ V, since f is surjective, we can write y as f(x) for some x ∈ X.  Then x ∈ f^–1(V), so y = f(x) ∈ f[f^–1(V)] ⊆ ↑f[f^–1(V)].  In the other direction, we wish to show ↑f[f^–1(V)] ⊆ V.  Since V is upwards-closed, it suffices to show f[f^–1(V)] ⊆ V.  Every point in the left-hand side if equal to f(x) for some x ∈ f^–1(V), and then f(x) ∈ V, which shows the claim.

If α plays according to σ’ and β plays according to τ on Y (the σ’/τ scenario), then we produce the following items:

• y₀, V₀ ≝ τ();
• U₀ ≝ ↑f[U’₀], where U’₀ ≝ σ(s(y₀), f^–1(V₀));
• y₁, V₁ ≝ τ(y₀, V₀, U₀);
• U₁ ≝ ↑f[U’₁], where U’₁ ≝ σ(s(y₀), f^–1(V₀), U’₀, s(y₁), f^–1(V₁)); note that U’₀ is obtained from the memory;
• etc.

If α plays according to σ and β plays according to τ’ on X (the σ/τ’ scenario), then we produce the following items:

• x₀, V’₀ where y₀, V₀ ≝ τ(), x₀ ≝ s(y₀) and V’₀ ≝ f^–1(V₀);
• U’₀ ≝ σ(x₀, V’₀); this is the same  U’₀ as in the σ’/τ scenario;
• x₁, V’₁ where y₁, V₁ ≝ τ(f(x₀), ↑f[V’₀], ↑f[U’₀]), x₁ ≝ s(y₁) and V’₁ ≝ f^–1(V₁); now  f(x₀)=f(s(y₀)),  ↑f[V’₀] = ↑f[f^–1(V₀)] = V₀ by (*), so defining  U₀ ≝ ↑f[U’₀], as in the σ’/τ scenario, y₁, V₁ is equal to τ(y₀, V₀, U₀),  as in the σ’/τ scenario;
• U’₁ ≝ σ(x₀, V’₀, U’₀, x₁, V’₁); since  x₀=s(y₀), V’₀=f^–1(V₀), x₁=s(y₁), and V’₁ = f^–1(V₁), this is the same U’₁ as in the σ’/τ scenario;
• etc.

In the end, we obtain the same sets U’_n and U_n = ↑f[U’_n] in both scenarios.

Then ∩_n ∈ N U’_n is non-empty, or the sets U’_n form a base of open neighborhoods of some point x_∞, or of some non-empty compact set Q’_∞, depending on whether σ is winning, convergent winning, or compactly winning.  By the same argument as in the stationary cases, ∩_n ∈ N U_n is non-empty, or the sets U_n form a base of open neighborhoods of some point x_∞, or of some non-empty compact set Q_∞ ≝ ∩_n ∈ N U_n which we prove equal to ↑f[Q’_∞], depending on whether σ is winning, convergent winning, or compactly winning.  Hence σ’ is winning, convergent winning, or compactly winning, depending on the case.  ☐

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— Jean Goubault-Larrecq (October 20th, 2025)