# A confusion about covering flatness

I'm reading this entry on nLab. But I'm confusing with the notion of covering-flatness. More precisely, I meet some trouble when I try to show that the $Sets$-valued flatness is a special case of covering-flatness.

The following is a possible proof.

Let $C$ be a small category and $F\colon C\to Sets$ a covering-flat functor. To show it is $Sets$-valued flat, consider an arbitrary finite diagram $D$ in the category $El(F)$ of elements of $F$. Then this diagram itself provides a cone $\alpha\colon\Delta_\ast\to FD'$ over the image of some diagram $D'$ in $C$ with same shape of $D$ under $F$. In this way, to give a cone over $D$ is the same to give a cone $\beta\colon\Delta_U\to D'$ over $D'$ such that $\alpha$ factors through the image of $\beta$ under $F$. Now, consider the limit cone $\lambda\colon\Delta_L\to FD'$ of $FD'$, then $\alpha$ factors through it, which gives rise to an element $a$ of $L:=\lim FD'$. Then by the describe of the sieve corresponding to the limit cone $\lambda$, there exists a cone $\beta\colon\Delta_U\to D'$ over $D'$ such that $F\circ\beta$ factors through $L$ and that the $a$ lies in the image of $U\to L$. Now, any preimage $u$ of $a$ in $U$ gives a cone $\Delta_{(u\colon\ast\to F(U))}\to D$ over $D$.

But the above argument seems failed if for instance the limit $L=\varnothing$.

What's more, consider the case that $F$ maps any object in $C$ to the empty set. Obviously it is not $Sets$-valued flat. However, consider that for any finite diagram $D\colon I\to C$, the only cone over $FD$ is induced by $\mathrm{id}_{\varnothing}$. The the sieve described in the statement corresponding to this cone is $\{\mathrm{id}_{\varnothing}\}$, which is a covering sieve in $Sets$. So, this $F$ is covering-flat but not $Sets$-valued flat. What's wrong?

• Can you clarify "Then this diagram [etc.]"? Specifically, what is $D'$ and where is this cone coming from? Oct 15 '15 at 23:01
• @ZhenLin, $D'$ is the composition of $D$ with the projection from $El(F)$ to $C$, which maps $(\ast\to F(U))$ to $U$. Let $D(i)=(s_i\colon\ast\to F(U_i))$ with transformation morphisms $D(i\to j)=\varphi_{ij}\colon U_i\to U_j$. Then there is a cone $\alpha$ given by $\alpha_i=s_i\colon\ast\to F(U_i)=D'(i)$. Oct 16 '15 at 7:27

Here's how I prefer to think of the definition of "covering-flat": a diagram $F : \mathcal{C} \to \mathcal{D}$ is covering-flat if and only if, for every finite diagram $X : \mathcal{I} \to \mathcal{C}$, the comparison morphism $F_! \varprojlim\nolimits_\mathcal{I} h_X \to \varprojlim\nolimits_\mathcal{I} h_{F X}$ in $[\mathcal{D}^\mathrm{op}, \mathbf{Set}]$ is a local epimorphism. (I assume $\mathcal{C}$ is small and $\mathcal{D}$ is locally small; if not, enlarge universe.)
In your "counterexample", $\mathcal{D} = \mathbf{Set}$ and $F = \emptyset$. Consider the case $\mathcal{I} = \emptyset$. Then $\varprojlim\nolimits_\mathcal{I} h_{F X} \cong 1$, whereas $F_! \varprojlim\nolimits_\mathcal{I} h_X \cong \varinjlim\nolimits_\mathcal{C} h_F$; but $(\varinjlim\nolimits_\mathcal{C} h_F) (K) = \emptyset$ for all non-empty $K$, so $F_! \varprojlim\nolimits_\mathcal{I} h_X \to \varprojlim\nolimits_\mathcal{I} h_{F X}$ is not a local epimorphism. Hence $F : \mathcal{C} \to \mathbf{Set}$ is not covering-flat.