Let $X$ be a topological space, and $F: \text{Set} \to \text{Cov}_X$ be the functor sending a set $S$ to the trivial covering $S \times X \to X$ of $X$ with fiber $S$.
I want to prove that $X$ is connected if and only if the functor $F$ is fully faithful.
I am not sure on how to proceed for the implication $F$ fully faithful $\Rightarrow$ $X$ connected. I thought that it would be useful to use the characterization of connected spaces saying that $X$ is connected $\Leftrightarrow$ every continuous function $f: X \to \{0,1 \}$ is constant.
So, I started by noticing that the functor sends a set-theoretic map $f: S \to T$ into $(f, \text{id}_X): S \times X \to T \times X$.
Consider $S=\{ \ast \}$ and $T=\{ 0,1\}.$ A generic continuous function $g: \{ \ast \} \times X \to \{0,1\}$ gives rise to a morphism of coverings $G: \{ \ast \} \times X \to \{0,1\} \times X$ by putting $G( \ast, x) = (g(\ast, x), x).$ Since the functor $F$ is fully faithful, every morphism of coverings can be written as $(g, \text{id}_X)$ for some $g: S \to T$, so, in our case, we get that $g( \ast, x) = g( \ast)$, i.e. $g: \{ \ast \} \times X \to \{0,1\}$ is actually constant.
I would like to use this reasoning to conclude that every continuous map $\{ \ast \} \times X \to \{0,1\}$ is constant, and, as a consequence, every continuous map $X \to \{0,1\}$ is constant, implying that $X$ is connected.
Does any of this make sense? Are the possible implications correct? Otherwise, I wanted to try with other characterizations of connectedness, but I got stuck. For example, I know that $Y \subseteq X$ is a clopen $\Leftrightarrow$ $Y \to X$ is a covering, but I didn't see a way of linking this fact to trivial coverings.
Any help will be highly appreciated. Thanks!
This is correct. In a bit more geometric language and slightly rephrased, your argument essentially goes like this: if $X$ is disconnected, we can write $X=U\cup V$ for disjoint opens $U$ and $V$ in $X$. Now, there is a (continuous) morphism of coverings $\{*\}\times X\to \{0,1\}\times X$ sending $(*,u)$ to $(0,u)$ for $u\in U$ and sending $(*,v)$ to $(1,v)$ for $v\in V$. This gives us already three different maps of coverings $\{*\}\times X\to \{0,1\}\times X$, the other two being of the form $(*,x)\mapsto (i,x)$ for $x\in X$, where $i=0,1$. But if $F$ is fully faithful, then only two such maps are allowed to exist, as there are only two maps of sets $\{*\}\to\{0,1\}$. This would then be a contradiction. (Again, this is just a restatement of your solution, but it sounds slightly more intuitive to me like this.)