$\pi_{\alpha} : X \to X_\alpha$ and $f_\alpha \in C(Y, X_\alpha) \Rightarrow \exists ! F \in C(Y,X_\alpha), f_\alpha = \pi_\alpha \circ F$.

Question

Question: If $\left\{ X_{\alpha} \right\}$ is a family of topological spaces, $X = \prod_{\alpha} X_\alpha$, with the product topology, is uniquely determined up to homeomorphism by the following property: There exists continuous maps $\pi_{\alpha} : X \to X_\alpha$ such that if $Y$ is any topological space and $f_\alpha \in C(Y, X_\alpha )$ for each $\alpha$, there is a unique $F \in C(Y,X)$ such that $f_{\alpha} = \pi_{\alpha} \circ F$.

My attempt:

Suppose there exists continuous maps $\pi_{\alpha} : X \to X_\alpha$ for each $\alpha$ in the index set. Take $Y$ to be a topological space and $f_\alpha \in C(Y, X_\alpha)$ for each $\alpha$. We will show there exists a unique $F \in C(Y,X)$ such that $f_\alpha = \pi_\alpha \circ F$.

Since $f_\alpha$ is continuous for some open $V \subseteq Y$ we have that $f_{\alpha}(V) = U \subseteq X_{\alpha}$. Then, by the continuity of $\pi_\alpha$ applying the inverse image we have; $\pi_\alpha^{-1}(U) = Q \subseteq X$ for some open set $Q$ of $X$. We claim that there exists a unique $F \in C(Y,X)$ such that $F(V) = Q$. We note that under the construction above that: $$f_\alpha \circ \pi^{-1}_\alpha(V) = \pi^{-1}_\alpha ( f_\alpha (V)) = \pi^{-1}_\alpha(U) = Q$$ Hence, we choose $F = f_\alpha \circ \pi^{-1}_\alpha$. For uniqueness, suppose $F(A) = F(B)$ for $A, B \in Y$, then, $\pi^{-1}_\alpha f_\alpha(A) = \pi^{-1}_\alpha f_\alpha(B)$, and by the continuity of this composition $A = B$.

Answer 1

So $X$ denotes the Cartesian product $\prod_\alpha X_{\alpha \in A}$, of the family of spaces $X_\alpha, \alpha \in A$ ,and $\pi_\alpha: X \rightarrow X_\alpha$ are the standard projections. $X$ gets the product topology, i.e. the one generated by the subbase $\mathcal{S} = \{\pi_\alpha^{-1}[O]; \alpha \in A, O \subseteq X_\alpha \text{ open }\}$. Note that by definition of continuity, this makes all $\pi_\alpha$ continuous maps, and it's the smallest topology to do so.

The first thing to be verified, is that indeed if $f_\alpha : Y \rightarrow X_\alpha$ are continuous maps, we can construct $F$ as required. As $\pi_\alpha \circ f$ should equal $f_\alpha$, it's clear what $F(y)$ should be (recall that $F(y)$ is a function from $A$ into $\cup_{\alpha \in A}X_\alpha$, so it's uniquely determined by the its values on $A$: $F(y)(\alpha) = (\pi_\alpha\circ F)(y) = f_\alpha(y)$, for each $\alpha \in A$. So unicity of $F$ is clear. The continuity of $f$ is also clear, as for any subbasic element $S = (\pi_\alpha)^{-1}[O] \in \mathcal{S}$, we have $$F^{-1}[S] = F^{-1}[(\pi_\alpha)^{-1}[O]] = (\pi_\alpha \circ F)^{-1}[O] = (f_\alpha)^{-1}[O]$$ which is open, as all $f_\alpha$ are continuous. What the question is about, is that this standard universal property actually determines the product $X$ up to homeomorphism.

So assume that the space $\hat{X}$ and the maps $p_\alpha: \hat{X} \rightarrow X_\alpha, \alpha \in A$ obey the same property for continuity into $\hat{X}$, as the standard product $X$ did with the $\pi_\alpha$. You then have to show that $X$ and $\hat{X}$ are homeomorphic. Well, we apply the property to the maps $p_\alpha$ and we get a unique $F: \hat{X} \rightarrow X$ such that $\pi_\alpha \circ F = p_\alpha$ for all $\alpha$.

Also applying the property to the $\pi_\alpha$ for $\hat{X}$ we get a unique map $G: X \rightarrow \hat{X}$ such that $p_\alpha \circ G = \pi_\alpha$ for all $\alpha$. Now consider $F \circ G: X \rightarrow X$, which obeys $$(F \circ G) \circ \pi_\alpha = F \circ (G \circ \pi_\alpha) = F \circ p_\alpha = \pi_\alpha$$

for all $\alpha$. Note that the identity on $X$ also satisfies this property for $X$ and the $\pi_\alpha$, and by the supposed unicity $F \circ G= \operatorname{id}_X$

a similar computation and unicity argument shows that likewise $G \circ F = \operatorname{id}_{\hat{X}}$. This shows that $F$ and $G$ are bijections and both homeomorphisms. So $\hat{X}$ is homeomorphic to $X$.