# A category with arbitrary products, but not all limits, or finite limits not commuting with filtered colimits?

I'm interested in finding an example of a locally small category $\mathcal{C}$ having

• small filtered colimits and

but lacking, either

1. all small limits,

or either

1. the property of finite limits commuting with filtered colimits.

If I'm not wrong, any decent forgetful functor $\mathcal{C} \longrightarrow \mathbf{Sets}$ from my category $\mathcal{C}$ to the category of sets, would make $\mathcal{C}$ fulfill properties (1) and (2). Hence, I have to discard all the usual concrete categories, haven't I?

I would be very grateful of any counterexamples (even the most obvious and embarrassing ones :-) ) anyone can tell me.

EDIT. Dear Zhen Lin and Hanno: thank you both for your examples. They are really useful to me.

• Have a look here. – Zhen Lin Dec 3 '14 at 11:25

1. The category of Kan complexes is a full subcategory of the category of simplicial sets that is closed under filtered colimits and small products, but it fails to have equalisers.

2. Consider the ordinal $\omega + 1 = \{ 0, 1, 2, \ldots, \omega \}$. Let $\mathcal{C}$ be the poset of $\sup$-closed subsets of $\omega + 1$. It is not hard to see that $\mathcal{C}$ is a complete lattice, so it is also cocomplete. Let $A = \{ 0, \omega \} \subset \omega + 1$ and let $B_n = \{ 0, \ldots, n \} \subset \omega + 1$. Clearly, the colimit of the chain $$B_0 \to B_1 \to B_2 \to B_3 \to \cdots$$ is just $\omega + 1$ itself. On the other hand, the colimit of $$A \cap B_0 \to A \cap B_1 \to A \cap B_2 \to A \cap B_3 \to \cdots$$ is $\{ 0 \}$. Thus, finite limits fail to commute with filtered colimits in $\mathcal{C}$.

Take a field $k$, put $R := k[x]/(x^2)$ and consider the full subcategory ${\mathscr C}$ of $R\text{-Mod}$ consisting of projective (=flat=injective, see below) $R$-modules.

• Directed limits of flat modules are always (i.e., for any ring) flat, so ${\mathscr C}$ inherits them from $R\text{-Mod}$.

• Products of injectives are always (i.e, in any category) injective, so ${\mathscr C}$ inherits them from $R\text{-Mod}$.

• $\alpha: R\xrightarrow{\cdot x} R$ does not have a kernel in ${\mathscr C}$:

First note that since $R\in{\mathscr C}$, a monomorphism in ${\mathscr C}$ is a monomorphism in $R\text{-Mod}$. In particular, a kernel of $\alpha$ in ${\mathscr C}$ would be a subobject of $R$ in $R\text{-Mod}$, of which there are only $\{0\}$, $k\cdot x$ and $R$. $\{0\}$ is excluded since $\alpha^2=0$ but $\alpha\neq 0$, so $\alpha$ is not a monomorphism in ${\mathscr C}$. Similarly, $R$ itself it excluded since $\alpha\neq 0$. Finally, $k\cdot x$ is excluded since it does not belong to ${\mathscr C}$.

Proof that flat, projective and injective modules coincide: Projective modules are flat, any flat module is a directed limit of finitely generated projective modules, and any directed limit of injective modules is injective (since $R$ is Noetherian). Hence, to show that $\text{Proj}\subset\text{Flat}\subset \text{Inj}$ it suffices to note that $R$ itself is injective. For $\text{Inj}\subset\text{Proj}$, take $v\in I\in\text{Inj}$ and suppose $x\cdot v=0$. Then, extending $k\cdot x\hookrightarrow I, x\mapsto v$, along the embedding $k\cdot x\hookrightarrow R$ shows that there exists some $v^{\prime}\in I$ with $x\cdot v^{\prime}=v$. Hence $X := \text{ker}(x\cdot -)=\text{im}(x\cdot -)\subset I$, so for $X^{\prime}$ a $k$-complement of $X$ in $I$, the canonical map $R\otimes_k X^{\prime}\to I$ is an isomorphism, so $I$ is projective.

Note that $\text{Proj}=\text{Inj}=\text{Flat}$ all consist of the "acyclic" $R$-modules, those where $\text{ker}(x\cdot -)=\text{im}(x\cdot -)$.