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

Question

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

• small filtered colimits and
• arbitrary small products

but lacking, either

1. all small limits,

or either

2. 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.

Answer 1

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 $\mathcal{C}$.

Answer 2

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 $\text{ker}(x\cdot -)=\text{im}(x\cdot -)$.