# 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?

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 $$P$$.

• Isn't your answer essentially the same as Malice Vidrine's? I know very little about sheaves, but it looks like you are using the same criterion, and reduce it to the same trivial case. – Arnaud D. Sep 21 '18 at 10:28
• @ArnaudD. Yes, it more or less is. I read through Malice Vidrine too quickly. – Derek Elkins left SE Sep 21 '18 at 17:05
• It's more succinctly presented than my answer, though :P – Malice Vidrine Sep 21 '18 at 20:01
• Sorry for being AWOL guys, was busy travelling and was trying to convince myself that the reformulation of the sheaf condition was true. I ended up adapting the proof of Yoneda's Lemma to show that $Nat(S,P)\cong P(A)$ due to the sheaf condition (and in the process, I think I understand what Kevin meant by a compatible family being determined by its component at the identity map of c), and then putting everything together to get the desired isomorphism. I think this proof works. Unfortunately, I can only accept 1 answer. Since this was the clearest one to me, I'm picking this one! – asldjk Sep 24 '18 at 15:14

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 $x$.

• Thanks! I'm still a bit confused, however. First, shouldn't the non-trivial maps $g$ have $cod(g)=dom(f)$ where $f\in M_c$ as opposed to having $cod(g)=c=cod(f)$? Second, I don't quite understand what you mean by a compatible family for $M_c$ being determined by its component $x$ at the identity map of $c$. I know we pick some $f\in S$, and we take the family of elts in $P(dom(f))$ satisfying some compatibility conditions - I'm confused as to how this family of elts has a component $x$, and where the identity map of $c$ fits in here. – asldjk Sep 21 '18 at 8:59

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.