Do homeomorphisms on a set uniquely determine a topology on the set? I am in the initial stage of a self study in algebraic topolgy and have a question that would appear trivial, except that I cannot come up with a answer:
On a given set, if we are given the quotient set of homeomorphically equivalent classes of elements, does this uniquely determine the topology on the larger set?
In other words, given the quotient set derived from some (unknown) topological equivalence on X, i.e., X/~ , is the implied topology on X unique? and if so, how can one induce it from X/~ ?
 A: You cannot in general reconstruct the topology on $X$. For example, take $X=\Bbb N$, let $\tau$ be the discrete topology on $X$, and let $\tau_2$ be the topology generated by the base $\mathscr{B}$ consisting of all sets of the form $\{2n,2n+1\}$ for $n\in\Bbb N$. Let $\sim$ be the equivalence relation whose equivalence classes are the members of $\mathscr{B}$. Then $X/\!\sim$ is a countable infinite discrete space both when $X$ is endowed with the topology $\tau$ and when $X$ is endowed with the topology $\tau_2$.
Less trivially, define $\sim$ on $\Bbb R$ by $x\sim y$ if and only if $x=y$, or $0\le x,y<1$. Then $\Bbb R/\!\sim$ is homeomorphic to $\Bbb R$ itself, which is the quotient of $\Bbb R$ by the identity relation.
A: Whenever you take a quotient by any equivalence relation, the map $X\rightarrow X/\sim$ is necessarily not injective (unless $\sim$ is trivial), and hence you lose information.
Whatever topology you choose to put on $X/\sim$, all you can tell is that preimages of open (resp. closed) subsets of $X/\sim$ are open (resp. closed) in $X$.
For example, if $X/\sim$ has the discrete topology, then you know that every fiber (ie, preimage of a point) is both open and closed, but you have control over the topology on $X$ restricted to that fiber. In fact, for any arbitrary choice of topology on every fiber of $X\rightarrow X/\sim$, the "union" of those topologies generates a topology on $X$ which induces the discrete topology on $X/\sim$ and restricts to your chosen topologies on every fiber.
