$\forall \alpha\in\mathscr{A},\; X_\alpha\cong Y_{\varphi(\alpha)}\implies \prod_\alpha X_\alpha\cong \prod_\beta Y_\beta$ (Homeomorphisms).

Question

The Unclear Claim

I'm having trouble understanding the following statement from Topology by James Dugundji chapter IV:

$\mathbf{2.6}$ Corollary. Let $\{X_\alpha\mid \alpha\in\mathscr{A}\}$ and $\{Y_\beta\mid \beta\in\mathscr{B}\}$ be two families of spaces, and $\varphi\colon \mathscr{A}\to\mathscr{B}$ a bijection. If for each $\alpha$, $X_\alpha\cong Y_{\varphi(\alpha)}$, then $\prod_\alpha X_\alpha\cong \prod_\beta Y_\beta$. In particular, $\prod_\alpha X_\alpha$ is unrestrictedly commutative.
Proof. This is immediate from $\mathbf{2.5}$ and III, $\mathbf{12.2}$.

$\mathbf{2.5}$ states

$\mathbf{2.5}$ Theorem. Let $\aleph(\mathscr{A})$ be arbitrary. For each $\alpha\in\mathscr{A}$, let $f_\alpha\colon X_\alpha\to Y_\alpha$ be a map. Define $\prod f_\alpha\colon\prod_\alpha X_\alpha\to\prod_\alpha Y_\alpha$ by $\{x_\alpha\}\to \{f_\alpha(x_\alpha)\}$. Then:

1. If each $f_\alpha$ is continuous, so also is $\prod f_\alpha$.
2. If each $f_\alpha$ is an open map, and all but finitely many are surjective, then $\prod f_\alpha$ is also an open map.

while III.$\mathbf{12.2}$ is just an elementary equivalence for homeomorphisms.

Why It's Unclear

I don't see how $\mathbf{2.5}$ implies $\mathbf{2.6}$. I feel like, for $\mathbf{2.6}$ to follow, it's conclusion should be $\prod_\alpha X_\alpha\cong \prod_\alpha Y_{\varphi(\alpha)}$ and not $\prod_\alpha X_\alpha\cong \prod_\beta Y_\beta$ since, what I think Dugundji is trying to say, is that the homeomorphisms from $X_\alpha$ to $Y_{\varphi(\alpha)}$ in $\mathbf{2.6}$ should be used as the $f_\alpha$'s on $\mathbf{2.5}$ to create the homeomorphism $\prod f_\alpha$. The thing is that, with this procedure, $\prod f_\alpha\colon \prod_\alpha X_\alpha\to \prod_\alpha Y_{\varphi(\alpha)}$ does not necessarily relate $\prod_{\alpha}X_\alpha$ to $\prod_\beta Y_\beta$.

What am I not seeing or understanding erroneously? Could someone provide the "immediate proof" of statement $\mathbf{2.6}$?

Answer 1

We need an extra ingredient, 2.5. is not sufficient to prove 2.6.

As Mehmet Kırdar writes in his answer, 2.5 shows that we get a homeomorphism $\Pi_{\alpha\in\mathscr A} f_{\alpha}: \Pi_{\alpha\in\mathscr A} X_\alpha\rightarrow \Pi_{\alpha\in\mathscr A} Y_{\varphi(\alpha)}$. It remains to show that $\Pi_{\alpha\in\mathscr A} Y_{\varphi(\alpha)} \approx \Pi_{\beta\in\mathscr B} Y_{\beta}$. By definition $$\prod_{\gamma \in \mathscr C} Z_\gamma = \left.\left\{ u : \mathscr C \to \bigcup_{\gamma \in \mathscr C} Z_\gamma \right| \forall \gamma \in \mathscr C: u(\gamma) \in Z_\gamma \right\} .$$ The product topology on $\prod_{\gamma \in \mathscr C} Z_\gamma $ is the coarsest topology such that all projections $p_\gamma : \prod_{\gamma \in \mathscr C} Z_\gamma \to Z_\gamma, p_\gamma(u) = u(\gamma)$, become continuous.

Now define $$\varphi^* : \Pi_{\beta\in\mathscr B} Y_{\beta} \to \Pi_{\alpha\in\mathscr A} Y_{\varphi(\alpha)}, \varphi^*(u) = u \circ \varphi .$$ Clearly $\bigcup_{\beta \in \mathscr B} Y_\beta = \bigcup_{\alpha \in \mathscr A} Y_{\varphi(\alpha)}$ and $(u \circ \varphi)(\alpha) = u(\varphi(\alpha)) \in Y_{\varphi(\alpha)}$, thus $\varphi^*$ is well-defined. It is moreover injective: If $\varphi^*(u) = \varphi^*(u')$, i.e. $u \circ \varphi = u' \circ \varphi$, we get $u = (u \circ \varphi) \circ \varphi^{-1} = (u' \circ \varphi) \circ \varphi^{-1} = u'$. It is also surjective: Each $v \in \Pi_{\alpha\in\mathscr A} Y_{\varphi(\alpha)}$ has the form $v = \varphi^*(v \circ \varphi^{-1})$ (note that $(v \circ \varphi^{-1})(\beta) = v(\varphi^{-1}(\beta)) \in Y_{\varphi(\varphi^{-1}(\beta))} =Y_\beta$ which means $v \circ \varphi^{-1} \in \Pi_{\beta\in\mathscr B} Y_{\beta}$ ).

Hence $\varphi^*$ is a bijection. The product toplogy on $\Pi_{\alpha\in\mathscr A} Y_{\varphi(\alpha)}$ is the coarsest topology such that all projections $p_\alpha : \Pi_{\alpha\in\mathscr A} Y_{\varphi(\alpha)} \to Y_{\varphi(\alpha)}$ become continuous. The bijection $\varphi^*$ induces a coarsest topology on $\Pi_{\beta\in\mathscr B} Y_{\beta}$ such that all functions $p_\alpha \circ \varphi^* : \Pi_{\beta\in\mathscr B} Y_{\beta} \to Y_{\varphi(\alpha)}$ become continuous. This topology makes $\varphi^*$ a homeomorphism. Since $\varphi$ is a bijection, the continuity of all $p_\alpha \circ \varphi^*$ is equivalent to the continuity of all $p_{\varphi^{-1}(\beta)} \circ \varphi^* : \Pi_{\beta\in\mathscr B} Y_{\beta} \to Y_{\varphi(\varphi^{-1}(\beta))} = Y_\beta$. But $p_{\varphi^{-1}(\beta)} \circ \varphi^*$ is nothing else than the projection $ \Pi_{\beta\in\mathscr B} Y_{\beta} \to Y_\beta$. Thus the above topology on $\Pi_{\beta\in\mathscr B} Y_{\beta}$ the product topology.

Answer 2

Too long for a comment:

For each $a\in\alpha$, we are given homeomorphisms $f_{a}: X_a\rightarrow Y_\varphi(a)$. So, we have a natural homeomorphism $\Pi_{a\in\alpha} f_{a}: \Pi_{a\in\alpha} X_a\rightarrow \Pi_{a\in\alpha} Y_\varphi(a)$ defined coordinatewise.

Therefore, all we have to do is to show that $\Pi_{a\in\alpha} Y_{\varphi(a)}\cong\Pi_{b\in\beta} Y_b$. This is also a natural homeomorhism defined by shuffling of coordinates. For finite products, it is easy to prove this. For example, $X\times Y\cong Y\times X$ by means of the map $(x,y)\rightarrow (y,x)$. But, here index sets are infinite. It is not easy to describe the shuffling map. We can define it tautologically $(y_{\varphi(a)})_\alpha\in \Pi_{a\in\alpha} Y_{\varphi(a)}$ must be sent to $(y_b)_\beta\in \Pi_{b\in\beta} Y_b$ so that $y_{\varphi(a)}=y_b$ iff $\varphi(a)=b$.

In both product topology and box topology both maps described above are homeomorphisms. Since, open sets are sent to open sets coordinatewise in both directions.

Answer 3

You are right, 2.6 is not immediate from 2.5.

2.5. allows to conclude that the product $\prod_\alpha f_\alpha : \prod_\alpha X_\alpha \to \prod_\alpha Y_\alpha$ of homeomorphisms $f_\alpha : X_\alpha \to Y_\alpha$ is a homeomorphism - but it is essential that a single index set is used. The situation in 2.6 is completely different: Dugundji works with a bijection $\varphi : \mathscr A \to \mathscr B$ and assumes that there exists a homeomorphism $f_\alpha : X_\alpha \to Y_{\varphi(\alpha)}$ for each $\alpha \in \mathscr A$. The task is to prove that we get a homeomorphism $$ \prod_\alpha X_\alpha \to \prod_\beta Y_\beta .$$

This requires a new approach. Let us do it a bit more general. Given indexed collections of spaces $(X_\alpha)_{\alpha \in \mathscr A}$ and $(Y_\beta)_{\beta \in \mathscr B}$ , consider a system $\mathfrak S$ consisting of

• a function $u : \mathscr B \to \mathscr A$
• for each $\beta \in \mathscr B$ a map $f_\beta : X_{u(\beta)} \to Y_\beta$.

Define
$$M_{\mathfrak S} : \prod_\alpha X_\alpha \to \prod_\beta Y_\beta, M_{\mathfrak S}((x_\alpha)_{\alpha \in \mathscr A}) = (f_{\beta}(x_{u(\beta)}))_{\beta \in \mathscr B} \tag{1}$$ This is a continuous map: Let $p_\alpha : \prod_\alpha X_\alpha \to X_\alpha$ and $q_\beta : \prod_\beta Y_\beta \to Y_\beta$ denote the projections. Then $q_\beta \circ M_{\mathfrak S} = f_\beta \circ p_{u(\beta)}$ and Theorems 2.1, 2.2 apply.

If $\mathscr B = \mathscr A$, $Y_\alpha = X_\alpha$, $u = id$ and all $f_\alpha = id$, then $\mathfrak S$ is the identity system $\mathfrak I $ for which clearly $M_{\mathfrak I} = id$.

Given one more indexed collection of spaces $(Z_\gamma)_{\gamma \in \mathscr C}$ and a system $\mathfrak T$ consisting of

• a function $v : \mathscr C \to \mathscr B$
• for each $\gamma \in \mathscr C$ a map $g_\gamma : Y_{v(\gamma)} \to Z_\gamma$

we define $$\mathfrak T \circ \mathfrak S = (u \circ v: \mathscr C \to \mathscr A, g_\gamma \circ f_{v(\gamma)} : X_{u(v((\gamma))} \to Z_\gamma) . \tag{2}$$ It is easily verified that $$M_{\mathfrak T \circ \mathfrak S} = M_{\mathfrak T} \circ M_{\mathfrak S} . \tag{3}$$

This can be applied as follows:

Let $u : \mathscr B \to \mathscr A$ be a bijection and the $f_\beta : X_{u(\beta)} \to Y_\beta$ be homeomorphisms. Consider the systems $\mathfrak S = (u, f_\beta)$ and $\mathfrak T = (u^{-1}, f^{-1}_{u^{-1}(\alpha)})$. We have $\mathfrak T \circ \mathfrak S = \mathfrak I$ and $\mathfrak S \circ \mathfrak T = \mathfrak I$, thus we can conclude that $M_{\mathfrak S}$ and $ M_{\mathfrak T}$ are homeomorphisms which are inverse to each other.

Answer 4

In your question you wrote: "I don't see how 2.5 implies 2.6."

Using the assumptions of Corollary 2.6, from Theorem 2.5 (and from properties of product of functions) you should have a map $h\colon \prod_\alpha X_\alpha \to \prod_\alpha Y_\alpha$ such that:

• $h$ is bijective,
• $h$ is continuous,
• $h$ is an open map.

These three properties already imply that $h$ is a homeomorphism. You could try to prove this yourself - but you can find a proof in various places. For example:

• Some questions on this site: $f$ is a homeomorphism iff $f$ is bijective, continuous and open or A bijective continuous map is a homeomorphism iff it is open, or equivalently, iff it is closed.
• ProofWiki: Bijection is Open iff Inverse is Continuous

A similar claim holds if you replace "open" by "closed". See also: Bijection is Open iff Closed (ProofWiki)