On the same set it is possible to have many different topologies. For example on $\mathbb R$ you have not only the usual topology, but also the discrete topology (all subsets are open) as the finest and the indescrete (only $\emptyset$ and $\mathbb R$ are open) as the coarsets possible topology - and many more.
Now if we fix a map $f\colon X\to Y$ and fix a topology on $Y$, but consider different topologies on $X$, then the continuity of $f$ depends on the topology on $X$. For example, with the discrete topology on $X$, all such maps $f$ are continuous; with the indiscrete it is much less "likely" that a given $f$ is continuous; with other topologies, the situation is inbetween: If $f$ is continuous wrt. to a coarser topology (e.g. wrt. to the indiscrete topology) the it is also continuous wrt. to this topology, whereas the converse need not hold.
$f^{-1}[Y]$ is the pre-image of a set under $f$, it's not about an inverse map. I prefer the notion with brackets over the one with braces for this notion.
If $f: X \to Y$ is any function and $B \subseteq Y$ then $f^{-1}[B] = \{x \in X \mid f(x) \in B\}$ is just the set of points of the domain that map into $B$. So if $B=Y$, by definition of a function from $X$ to $Y$, all $f(x) \in Y$ and so $f^{-1}[Y]=X$ follows.
So if $f$ is constant with value $c$, $f^{-1}[B]=X$ if $c \in B$ and $\emptyset$ otherwise. So always open in $X$ regardless of what topology $X$ or $Y$ have.
As an afterhought: suppose that $X$ has the trivial topology $\{\emptyset,X\}$, and if $f$ is not constant so there are $x,x'$ in $X$ such that $f(x) \neq f(x')$. If in $Y$ there is an open set $O$ that contains just one of the points $\{f(x), f(x')\}$ then $f$ cannot be continuous as $f^{-1}[O]$ is non-empty (it contains $x$ or $x'$) and also not equal to $X$ (it misses $x$ or $x'$; the other one), so $f^{-1}[O]$ cannot be open. But e.g. $X=\{1,2,3\}$ in the trivial topology to $Y=\{1,2,3\}$ in the topology $\{\emptyset, \{2,3\}, Y\}$ has a non-constant continuous function (defined by $f(1)=2, f(2)=3, f(3)=2$, which has $f^{-1}[\emptyset] = \emptyset, f^{-1}[Y] =X, f^{-1}[\{2,3\}] = X$ so is continuous)).
So it's not true that a trivial topology only has constant maps to non-trivial $Y$, it will depend on the topology of $Y$.
In fact, let $I(2)$ be the space $\{0,1\}$ in the indiscrete/ trivial topology.
$Y$ is $T_0$ iff every continuous $f: I(2) \to Y$ is constant
Proof: If $Y$ is $T_0$ and $f$ is a function from $I(2)$ to $Y$ that is non-constant then $f(0) \neq f(1)$ and so by $T_0$-ness there is an open set $O$ in $Y$ such that $\{f(0),f(1)\} \cap O$ has one member, say $f(0)$ WLOG. Then $f^{-1}[O]=\{0\}$ is not open in $I(2)$. So $f$ is not continuous.
If we assume the mapping property, we can show that $Y$ is $T_0$: suppose $Y$ is not $T_0$. Then there are two points $y_1\neq y_2$ in $Y$ such that for every open set $O$ in $Y$ we have $y_1 \in O \land y_2 \in O$ or $y_1 \notin O \land y_2 \notin O$; we cannot separate them by an open set. But that implies that the map $f$ sending $0$ to $y_1$ and $1$ to $y_2$ is continuous, as $f^{-1}[O]=\emptyset$ or $=I(2)$ depending on which is the case for $O$. And $f$ is non-constant, so that would contradict the mapping property. So $Y$ is $T_0$.
So in topological category theory we could say that the $T_0$ spaces are the $I(2)$-co-connected spaces in $\mathbf{Top}$, or some such term. Connected spaces are $D(2)$-connected (where $D(2)=\{0,1\}$ in the discrete topology), as is classical (all continuous functions into $D(2)$ are constant).
Best Answer
Every function f from the empty space E into any other space S
is continuous because f$^{-1}$(U) for any set U is empty.
On the other hand the only function into E is the empty function.
Thus every function from E or into E is continuous.
E is the only space with exactly one open set.
Is E T$_0$, T$_1$, Hausdorff, regular, etc.?
Is E connected, disconnected, discrete, metrizable?