A space whose powers "generate" all spaces via quotients I've been asked to prove (or disprove) that there exists a topological space $X$ with the following property:

For every space $T$ there is a set $I$ and a homeomorphism $T\cong A$ where $A$ is a quotient of a subspace of $X^I = \prod_{i\in I} X$ (product topology)

Of course, this smells like a universal property, and a rather strong one, so I'm inclined to think there is no such $X$. But how can one prove that such a space does not exist?
 A: You can do this without even needing quotients.  Let $X=\{0,0',1\}$, with the topology $\{X,\emptyset,\{1\}\}$.  Given any space $T$, let $I$ be the set of all continuous maps $T\to X$.  There is a canonical continuous map $F:T\to X^I$ given by $F(t)(f)=f(t)$.  I claim that $F$ is a homeomorphism onto its image, so $T$ is homeomorphic to the subspace $A=F(T)\subseteq X^I$.
First, $F$ is injective, since the subspace $\{0,0'\}$ of $X$ is indiscrete and so any map $T\to \{0,0'\}$ is continuous.  In particular, given any two distinct points of $T$, we can find a continuous map $T\to X$ which sends one of them to $0$ and the other to $0'$.
Second, $F$ is open as a map onto its image.  Indeed, let $U\subseteq T$ be open.  Then the function $f:T\to X$ sending $U$ to $1$ and $T\setminus U$ to $0$ is continuous.  The image $F(U)$ is then exactly the set of elements of $F(T)$ whose $f$ coordinate is $1$.  This is an open subset of $F(T)$ since $\{1\}$ is open in $X$.  Thus $F$ is open as a map to $F(T)$.
A: Summarising the other answers:
The embedding theorem shows that any topological space $S$ is homeomorphic to a subspace of $X^I$ for some index set $I$, where $X = \{0,1,2\}$ with topology $\{\{0\}, \emptyset, X\}$ (a sort of "fat Sierpinski space"). See Wofsey's answer for some more details. The essence is that $X$ allows us to separate points and separate points and closed sets by continuous functions into $X$.
This shows that the variant in the "quotients of subspaces" is answered by yes, without taking quotients at all; not very interesting.
The variant "quotients of $X^I$ " is answered by user 87690's answer:
Suppose $X$ existed. Let $S$ be the discrete space on $|X|^+$ (successor cardinal) points, and suppose we have a surjective quotient map $q: X^I \to S$, for some index set $I$.  
Indeed standard theorems show that if $\kappa$ is the smallest infinite cardinality of a dense subset of $X$, so $\kappa \le |X|$, then $X^I$ has no pairwise disjoint family of open non-empty sets of size $>\kappa$. But this is contradicted by $\{q^{-1}[\{s\}], s \in S\}$. This contradiction shows that $X$ cannot exist for all topological spaces $S$.
IMHO this last answer is more interesting as it does use quotients, albeit in the guise of just a continuous image.
A true dual to the powers/subspace variant would be:

Does there exist a space $X$, such that for any topological space $S$, there is a (surjective) quotient map $q: \sum_{i \in I} X_i \to S$ where all $X_i = X$, and $\sum$ denotes disjoint sums of spaces in their sum topology. 

(this idea occurs in the theory of sequential spaces, where we take quotients of copies of a convergent sequence); I think it might hold. [EDIT] No such luck, by the same argument  that sequential spaces are countably tight, we get that the tightness of quotients of sums of copies of $X$ will not exceed $t(X)$... Thx to the comment.
A: Note that there is no space $X$ such that every space $T$ is homeomorphic to some quotient $A$ of a whole power $X^I$ (rather than of some of its subspaces).
By $Δ$-system lemma, $c(∏_{i ∈ I} X_i) = \sup\{c(∏_{i ∈ F} X_i) : F ⊆ I \text{ finite}\}$. Hence, we have $c(A) ≤ c(X^I) ≤ \sup_{n ∈ ω} c(X^n) ≤ d(X)$. Therefore a space $X$ of density $κ$ can generate only spaces $A$ of cellularity $≤ κ$.
