I've just started learning about toposes and I have a stupid question to ask.
Suppose we are given a small category $\mathcal{C}$ with trivial topology $T$ on it, (where the trivial topology $T$ on a category $\mathcal{C}$ is the Grothendieck topology on $\mathcal{C}$ whose only covering sieves are the maximal ones, i.e. $M_c:=\{f| cod(f)=c\}$.
Question: Why is it that the $T$-sheaves on $\mathcal{C}$ are just the presheaves on $\mathcal{C}$?
By definition, a $T$-sheaf is a presheaf $P:\mathcal{C}^{op}\rightarrow Set$ on $\mathcal{C}$ such that for every $M_c$ and every family $\{x_f\in P(dom(f))| f\in S\}$ whereby $P(g)(x_f)=x_{f\circ g}$ given any $f\in S$ and any arrow $g$ in $\mathcal{C}$ composable with $f$, there exists a unique elt $x\in P(c)$ such that $x_f=P(f)(x)$ for all $f\in S$. I think I more or less understand this definition, and how it is basically a categorical formulation of the gluing axiom from sheaf theory.
However, I'm struggling to see how the fact that we are dealing with the trivial topology implies that every presheaf is a sheaf, i.e. that we can obtain this unique element $x\in P(c)$. I was trying to use the functoriality of the presheaves $P$, but that didn't seem to get me much. Can somebody give me a hint please?
As yet another take, I like the following.
First, it is a good (and useful!) exercise to show that the sheaf condition on a site $(\mathcal C,J)$ is equivalent to the following statement: For every object, $A$, of $\mathcal C$, and covering sieve $S$ on $A$ (i.e. a subfunctor of $\mathsf{Hom}(-,A)$ contained in $J(A)$), a presheaf $P$ is a sheaf if and only if $$P(A)\cong\mathsf{Nat}(\mathsf{Hom}(-,A),P)\cong \mathsf{Nat}(S,P)$$ i.e. the arrow $\mathsf{Nat}(\mathsf{Hom}(-,A),P)\cong\mathsf{Nat}(S,P)$ via precomposition by $S\rightarrowtail\mathsf{Hom}(-,A)$ is invertible. (The first isomorphism is just Yoneda.)
Clearly, if for each $A$ the only covering sieve on $A$ is $\mathsf{Hom}(-,A)$ itself, then the above form of the sheaf condition holds trivially for every presheaf $P$.