The empty set is covered by an empty family of sets. Therefore any two sections of F(empty set) are vacuously equal, so the F(empty set) can be at most a singleton. Now, why does this mean that it is in fact the terminal object, as Vakil claims in his Algebraic geometry notes? In Sets, this is true, but why in general? In particular, what happens if we are talking about a sheaf of rings with unity. Then I think that we actually have a problem because rings with unity have at least two elements.

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    $\begingroup$ Rings with unity can have one element if $1 = 0$. Never forget trivial cases :D $\endgroup$ Oct 24, 2013 at 18:13

2 Answers 2


For any index set $I$ we can consider the covering of $U=\emptyset$ by the sets $U_i=\emptyset$ for $i\in I$. This gives us that $F(U)$ in $$ \tag1F(U)\to\prod_{i\in I}F(U_i){{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}} \prod_{i,j\in I}F(U_i\cap U_j)$$ is an equalizer. For $U=\emptyset$ we may in fact pick $I=\emptyset$ and this still works. But the empty product of whatever is terminal, so $(1)$ becomes $$ F(\emptyset)\to T{{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}}T$$ where $T$ is terminal and the two arrows on the right are $\operatorname{id}_T$. The equalizer condition says that for any object $X$ together with a morphism $X\to T$ such that in $X\to T{{{} \atop \longrightarrow}\atop{\longrightarrow \atop {}}}T$ both paths to the right $T$ give the same morphism, there exists a unique morphism $X\to F(\emptyset)$ that makes $$\tag2 \begin{matrix}F(\emptyset)&\longrightarrow&T\\\uparrow&\nearrow\\X\end{matrix}$$ commute. But the condition about the morphism $X\to T$ is not really a condition: There exists a unique morphism from $X$ to $T$ any way and the equality of the two paths is always given (again by uniqueness of morphism to $T$). On the other hand, any morphism $X\to F(\emptyset)$ make $(2)$ commutative. Therefore $F(\emptyset)$ is simply an object such for any $X$ there exists a unique morphism $X\to F(\emptyset)$. - In other words, $F(\emptyset)$ is terminal.

Now that $(1)$ can be used to define the notion of sheaf only for a category with products including the empty product, hence also having terminal objects. The category Ring of rings with unity has products only for the nonempty case, so $(1)$ does not apply. More precisely, we can use $(1)$ for Ring valued sheaves only if $I\ne \emptyset$. Still, the formulation of locality with sections also postulates that $F(\emptyset)$ has only one element, which is impossible. Everything makes sense again if we augment Ring by adding the zero ring artificially as terminal object.

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    $\begingroup$ Why do you think that $\textbf{Ring}$ has no terminal object? The zero ring is terminal (and it has a unit, $0=1$). It is not artificial to add the zero ring to $\textbf{Ring} \setminus \{0\}$ (which you call $\textbf{Ring}$), but rather to define rings (with unit or not, doesn't matter) to be non-trivial (unfortunately several authors and lecturers do this!). Because then in almost every construction in ring theory you have to make case distinctions ... The work on universal algebra is 100 years old, still neglected ... $\endgroup$ Oct 24, 2013 at 23:11
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    $\begingroup$ Actually you prove it yourself (sheaves on empty set) that only considering non-trivial rings is artificial and causes case distinctions .... $\endgroup$ Oct 25, 2013 at 11:34
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    $\begingroup$ Hagen, I wanted to thank you before for your awesome answer, but since they discourage it here I remained silent. But I just found a minor typo in your answer that I could not correct myself (because it is shorter than 6 characters) and will vindicate this post. It should read "The equalizer condition says that for any object X together with a morphism X to T" and not "O to T". $\endgroup$
    – Rodrigo
    Nov 1, 2013 at 3:46

If $F$ is a sheaf valued in a category $C$ with terminal object $t$, then the empty covering of $\emptyset$ shows that $F(\emptyset) \to t \rightrightarrows t$ is an equalizer. Since $t$ is terminal, the two morphisms $t \rightrightarrows t$ agree, hence $F(\emptyset) \to t$ is an isomorphism.

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    $\begingroup$ Does this work in any category? $\endgroup$ Mar 12, 2014 at 12:21
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    $\begingroup$ Do you mean if we don't need the existence of a terminal object? How do you want to formulate the sheaf axiom without products? $\endgroup$ Mar 12, 2014 at 14:05
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    $\begingroup$ No I didn't mean that. I can see this for concrete categories with terminal object in which bijective morphisms are isomorphism. Can you explain your answer a bit further? $\endgroup$ Mar 12, 2014 at 16:27
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    $\begingroup$ I don't know what I should explain more since every step is explained + trivial. $\endgroup$ Mar 12, 2014 at 22:13
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    $\begingroup$ I learned sheaf axioms by sections which are elements. I guess you are using a generalized definition of sheaf possibly for nonconcrete categories? I apologize for commenting many times. I think I get what you are talking about. $\endgroup$ Mar 13, 2014 at 1:04

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