Reconstructing relations with the image relation of a topology For any topological space $(X,\tau)$ we define $$R_{im}(X,\tau) := \{(x,y)\in X^2: (\exists f:X\to X) \text{ continuous and surjective with } f(x) = y\}.$$
Clearly, $R_{im}(X,\tau)$ is reflexive. This relation is also transitive because the composition of two continuous surjective maps is continuous and surjective.
Given a non-empty set $X$ and a reflexive and transitive relation $R$, is there a topology $\tau_R$ on $X$ such that $R_{im}(X,\tau_R) = R$? If not, is the answer positive, if we require $R$ to be
a) an equivalence relation;
b) an ordering relation?
 A: The answer is no to the general question, and also to question (b), for the following simple reason (which works whether or not the space is finite): if $f:X\to X$ is surjective and $f(x)=y$ for some $x\neq y$, then there must be some $w\neq x$ with $f(w)=x$. Thus, the relation $R_{im}$ must have the property that whenever $x\mathrel{R_{im}} y$ and $x\neq y$, then there must be some $w\neq x$ with $w\mathrel{R_{im}} x$. In particular, $R_{im}$ can have no minimal elements, and so there are numerous counterexamples to the main question and question (b).
Update. Meanwhile, the answer to (a) is positive for finite sets, as asked in the comments:
Theorem. Every equivalence relation on a finite set $X$ arises
as $R_{im}(X,\tau)$ for some topology $\tau$ on $X$.
Proof. Suppose that $R$ is an equivalence relation on a finite
set $X$. We may place a linear pre-order $\trianglelefteq$ on $X$ in such
a way that $x\mathrel{R} y$ is equivalent to $x\trianglelefteq
y\trianglelefteq x$. Define the topology $\tau$ to have as open
sets exactly the upward closed sets, which have the form
$U_x=\{y\in X\mid x\trianglelefteq y\}$, plus the empty set. (This
collection is closed under arbitrary unions and intersections.)
I claim that $R_{im}(X,\tau)=R$. First of all, the topology does
not distinguish between points within any $R$-equivalence class,
and so we may permute within them at will. Thus, $R\subset
R_{im}(X,\tau)$. Conversely, suppose that $f:X\to X$ is continuous
and surjective. I claim that $f$ is merely permuting within each
$R$-class. First, note that $f$ is $\trianglelefteq$-preserving:
if $f(x)=y$, then $x$ is in the preimage $f^{-1}(U_y)$, which is
open, and so if $x\trianglelefteq x'$, then since $x'$ is in any
open set that $x$ is in, it follows that $y\trianglelefteq f(x')$,
and so we have $x\trianglelefteq x'\longrightarrow
f(x)\trianglelefteq f(x')$. From this, it follows that $f$ must
take each $R$-equivalence class into a single $R$-equivalence
class. Since $X$ is finite and $f$ is surjective, we know that $f$
is a permutation of $X$, and so $f$ must take the least $R$-class
to itself, and the next and so on, since otherwise we'd violate
the $\trianglelefteq$-preserving property. In other words, $f$ is
merely permuting the points inside each equivalence class, and so
$R_{im}(X,\tau)\subset R$ and hence $R_{im}(X,\tau)=R$, as
desired. QED
