Presheaves are sheaves for the trivial topology 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? 
 A: A compatible family for the maximal sieve $M_c$ is uniquely determined by its component $x$ at the identity map of $c$, using the compatibility condition with every nontrivial map $g:c’\to c$ to determine the other components. So the desired element of $P(C)$ is just $x$.
A: 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 $P$. 
A: Another way that the sheaf condition on a site can be described is as an extension property. A covering family on an object of a site is just a sieve (or a family of morphisms that generates one), and a sieve is just a subfunctor of a representable functor. So let $(\mathcal{C},T)$ be a site; the sheaf condition says that a presheaf $F:\mathcal{C}^{op}\to\mathbf{Set}$ is a sheaf with respect to $T$ if for any object $A$ of $\mathcal{C}$ and sub-representable functor $R\in T(A)$, every natural transformation $\alpha: R\to F$ extends uniquely to a natural transformation $\bar{\alpha}:\mathcal{C}(-,A)\to F$. It's easy to see that such an $\alpha$ just picks out a matching family, and an extension $\bar{\alpha}$ is a collation of that family.
So if for all $A$ we just have $T(A)=\{\mathcal{C}(-,A)\}$, the only families we're requiring to be collatable are the ones of the form $(F(f)(x))_{f:*\to A}$ for some $x\in F(A)$, which isn't a special condition at all; such families are already collated, because a natural transformation $\mathcal{C}(-,A)\to F$ will always trivially extend to itself in a unique way.
