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I was just working through some exercises in "Categories and Sheaves" by Kashiwara-Schapira, and encountered some problems with exercise 3.6 , on pg. 91.


Notation / Terminology

  • $C^\wedge $ denotes the category of presheaves on C, i.e functors $C^{op}$ to Set.

  • A category is idempotent complete if every morphism $q:X \rightarrow X$ satisfying $$q^2= q $$ factors as $q=g \circ f$ with $f:X \rightarrow Y$ an epimorphism and $g:Y \rightarrow X$ a monomorphism.

  • A functor $F:C \rightarrow C'$ is left-exact if the following category is co-filtered: The category $C^U$, with objects morphisms $f:U \rightarrow F(X)$ for $X \in C$, $f \in Hom_C(U,F(X)$ and morphisms between $f:U \rightarrow F(X)$ and $g:U \rightarrow F(Y)$ consists of $h \in Hom_C(X,Y)$ such that $g = F(h) \circ f$.

Problem

Assume that C is idempotent complete. Prove that the Yoneda Functor $h_c:C \rightarrow C^\wedge$ is left-exact if andonly if C admits finite projective limits.

OK, my strategy is to prove that C has a terminal object, finite products, and kernels. I have managed to prove that it has a terminal object, by considering the terminal functor of $C^op \rightarrow Set$. However, I haven't been able to show that it admits product, or kernels. Anyone got any ideas on how this can be done? Any hint, solutions, whatever are welcome. Thanks!

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Let me first note that what Kashiwara and Schapira call "left exact" is a considerably stronger condition than what most people think of, namely, the preservation of finite limits. The Yoneda embedding is always left exact in this weaker sense – in fact, it preserves all limits than exist in $\mathcal{C}$. (This is an easy exercise and amounts to unwinding definitions.) We will need to make use of this fact.

For clarity, let me call a functor that is left exact in the strong sense "representably flat".

Theorem. Suppose all idempotents in $\mathcal{C}$ split. Then, the Yoneda embedding is representably flat if and only if $\mathcal{C}$ has all finite limits.

Proof. First, assume $\mathcal{C}$ has all finite limits. Let $P$ be any presheaf on $\mathcal{C}$. Recalling that a category with all finite limits is automatically cofiltered, to show that $(P \downarrow H_\bullet)$ is cofiltered, it is enough to show that it has all finite limits. But $H_\bullet$ preserves finite limits and the forgetful functor $(P \downarrow H_\bullet) \to \hat{\mathcal{C}}$ creates them, thus, $(P \downarrow H_\bullet)$ is indeed cofiltered. (This can also be shown by hand using elementary methods.)

Conversely, suppose the Yoneda embedding is representably flat. To show that $\mathcal{C}$ has a terminal object, consider the cofiltered category $(1 \downarrow H_\bullet)$. It is non-empty, so there is a morphism $1 \to H_c$. By unwinding definitions, this means we have a morphism $f_d : d \to c$ for each object $d$ in $\mathcal{C}$ such that $f_d \circ k = f_{d'}$ for all morphisms $k : d' \to d$ in $\mathcal{C}$. In particular, $f_c : c \to c$ must be idempotent and splits as $f_c = s \circ r$ for some $r : c \to e$, $s : e \to c$ such that $r \circ s = \textrm{id}_e$. Notice that we have a morphism $d \to e$ for any $d$, namely $r \circ f_d$. But for any other $g : d \to e$, we must have $$f_d = f_c \circ s \circ g = s \circ r \circ s \circ g = s \circ g$$ and therefore $r \circ f_d = g$ for all $g : d \to e$; hence $e$ is a terminal object of $\mathcal{C}$.

Now, let $x$ and $y$ be any two objects of $\mathcal{C}$. To show that $x \times y$ exists in $\mathcal{C}$, we consider the cofiltered category $(H_x \times H_y \downarrow H_\bullet)$. We already know this is a non-empty category because we have the two projections $\pi_1 : H_x \times H_y \to H_x$ and $\pi_2 : H_x \times H_y \to H_y$, but since it is cofiltered, we also get a morphism $f : H_x \times H_y \to H_c$ and morphisms $p_1 : c \to x$, $p_2 : c \to y$ such that $H_{p_1} \circ f = \pi_1$ and $H_{p_2} \circ f = \pi_2$. Unwinding definitions, this means for every pair $g : d \to x$, $h : d \to y$, there is a morphism $f(g, h) : d \to c$ such that $p_1 \circ f(g, h) = g$ and $p_2 \circ f(g, h) = h$. Moreover, for any $k : d' \to d$, we have $f(g, h) \circ k = f(g \circ k, h \circ k)$. In particular, $$f(p_1, p_2) \circ f(p_1, p_2) = f(p_1 \circ f(p_1, p_2), p_2 \circ f(p_1, p_2)) = f(p_1, p_2)$$ so $f(p_1, p_2) : c \to c$ is idempotent. Suppose $f(p_1, p_2) = s \circ r$ is a splitting, where $r : c \to e$ satisfies $r \circ s = \textrm{id}_e$. I claim $e$ is the product of $x$ and $y$ in $\mathcal{C}$, with projections given by $p_1 \circ s$ and $p_2 \circ s$. Indeed, suppose $\ell : d \to e$ is any morphism such that $p_1 \circ s \circ \ell = g$ and $p_2 \circ s \circ \ell = h$. Then, $$r \circ f(g, h) = r \circ f(p_1 \circ s \circ \ell, p_2 \circ s \circ \ell) = r \circ f(p_1, p_2) \circ s \circ \ell = r \circ s \circ r \circ s \circ \ell = \ell$$ as required for a product.

Finally, let $g, h : x \to y$ be any two morphisms of $\mathcal{C}$. To show that the equaliser of $g$ and $g$ exists in $\mathcal{C}$, we consider the cofiltered category $(E \downarrow H_\bullet)$, where $E$ is the equaliser of $H_g, H_h : H_x \to H_y$ in $\mathcal{C}$. Since the category is cofiltered, there exists a morphism $f : E \to H_c$ and a morphism $i : c \to x$ such that $H_i \circ f$ is the canonical inclusion $E \to H_x$ and $g \circ i = h \circ i$. Unwinding definitions, this means for any two morphisms $j : d \to x$ such that $g \circ j = h \circ j$, there exists a morphism $f(j) : d \to c$ such that $i \circ f(j) = j$, and for any morphism $k : d' \to d$, we have $f(j \circ k) = f(j) \circ k$. Therefore, $$f(i) \circ f(i) = f(i \circ f(i)) = f(i)$$ and we can split $f(i)$ as $s \circ r$ for some split epimorphism $r : c \to e$. By this point it should be clear that $e$ is the equaliser of $g$ and $h$ in $\mathcal{C}$. Let us check that it works. Given any $\ell : d \to e$ such that $i \circ s \circ \ell = j$, we must have $$r \circ f(j) = r \circ f(i \circ s \circ \ell) = r \circ f(i) \circ s \circ \ell = \ell$$ and so $e$ is indeed the equaliser of $g$ and $h$, with canonical inclusion $i \circ s$. ◼

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Thank you! Wonderful. –  Dedalus Aug 5 '12 at 8:49

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