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According to Wikipedia article

"Unlike for algebraic objects, the inverse limit might not exist in an arbitrary category."

But when constructing the inverse limit of $(M_\alpha)_\alpha$ as a coherent subset of $\prod_{\alpha}M_\alpha$ and checking if it satisfies the universal property, the algebraic structure doesn't seem necessary. Is it referring to more general category where the objects are not even sets?

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Or categories where the product object does not exist. Take the category of finite groups and consider the same inverse system that defines the $p$-adic integers: the product object does not exist. – Arturo Magidin Oct 17 '11 at 2:49
Take a category with precisely two objects and two morphisms i.e. the two identity morphisms). – Pierre-Yves Gaillard Oct 17 '11 at 3:20
@Pierre-YvesGaillard: But wouldn't the only inverse system there be the constant system? It seems to me that in that situation, the inverse limit would just be the object. – Arturo Magidin Oct 17 '11 at 3:47
Dear @Arturo: I hope the following is correct. Assume that the distinct objects are $A$ and $B$, that the morphisms are $1_A$ and $1_B$, and that there are no other objects and no other morphisms. Then the empty product doesn't exist (it would be a terminal object), and $A\times B$ doesn't exist either (it would map to $A$ and to $B$). – Pierre-Yves Gaillard Oct 17 '11 at 4:35
@Pierre-YvesGaillard: Yes, the product of $A$ and $B$ does not exist (though it would be an initial object, I think since it has maps to $A$ and to $B$). But in this category, any inverse system does have an inverse limit, because the only inverse systems are the trivial ones. So while the construction using products does not work, inverse limits still exist in that category... – Arturo Magidin Oct 17 '11 at 4:43
up vote 4 down vote accepted

If a category $\mathcal{C}$ has all small products (that is, any family indexed by a "small" set has a product), and has equalizers, then every small limit in $\mathcal{C}$ exists, and in particular every inversely directed system of objects of $\mathcal{C}$ has an inverse limit; the construction is essentially the one from $\mathcal{S}et$ (and other categories that have a forgetful functor with an adjoint).

Explicitly, here is Proposition 7.6.6 from George Bergman's An Invitation to General Algebra and Universal Constructions:

Proposition. Let $\mathbf{C}$ be a category and $\mathbf{D}$ a small category, and let $\alpha$ be an infinite cardinal such that $\mathbf{D}$ has fewer than $\alpha$ objects and fewer than $\alpha$ morphisms. Then, if $\mathbf{C}$ has products of all families of fewer than $\alpha$ objects, and has equalizers, then every functor $\mathbf{D}\to\mathbf{C}$ has a limit.

(You can think of inverse limits as the categorical limit of a functor from the category of an inversely directed set, viewed as a category, to $\mathbf{C}$.)

The construction mimics the one from $\mathcal{S}et$: let $F\colon\mathbf{D}\to\mathbf{C}$ be a functor.

$$P = \prod_{X\in\mathrm{Ob}(\mathbf{D})} F(X)\qquad\text{and}\qquad P' = \prod_{\stackrel{X,Y\in\mathrm{Ob}(\mathbf{D})}{f\in\mathbf{D}(X,Y)}} F(Y).$$ Let $\pi_X$ be the projection morphisms from $P$ to $F(X)$, and $p_{X,Y,f}$ the projection morphisms from $P'$ to $F(Y)$.

Define $a,b\colon P\to P'$ as follows: since they are maps into a product, we can specify the value of the compositions $p_{X,Y,f}a$ and $p_{X,Y,f}b$, which will uniquely determine $a$ and $b$. First, $a$ is the map that satisfies $p_{X,Y,f}a = \pi_Y$ (that is, project $P$ onto $F(Y)$, then map $F(Y)$ to $F(Y)$ via the identity; do this for each triple $(X,Y,f)$; this gives a map from $P$ to $P'$). Second, $b$ is the map that satisfies $p_{X,Y,f}b = F(f)\pi_X$ (that is, project $P$ onto $F(X)$, then map from $F(X)$ to $F(Y)$ via $F(f)$).

Let $k\colon L\to P$ be the equalizer of $a$ and $b$. The universal property of $L$ and $k$ is equivalent to the property that the maps $\pi_Xk\colon L\to F(x)$ give commuting diagrams with the morphisms $F(f)$ and are universal. The object $L$ and the morphisms $\pi_Xk$ satisfy the universal property characterizing $\lim\limits_{\leftarrow} F$, so $L$ is a limit for $F$. $\Box$

(There is a dual result for colimits, when the category $\mathbf{C}$ has all sufficiently large coproducts and has coequalizers).

In fact, this proposition captures exactly how much structure you need to be able to carry out the construction from $\mathcal{S}et$/$\mathcal{G}roup$/etc in an arbitrary category: you need to be able to construct the product (which for arbitrary categories need not exist), and you must be able to specify the "compatibility conditions" in some way, which turns out to be equivalent to the existence of a suitable equalizer. If there is no equalizer, you may not be able to "specify" the appropriate "subobject" of the product.

So... you may be able to carry out the relevant construction even in categories that are not concrete (the objects need not be sets, and the arrows need not be set-theoretic functions between the objects), if you have sufficient structure. Or you may have a concrete category (such as "finite groups"), but where there isn't sufficient structure to carry out the construction.

The reason that you don't need to check the "relevant algebraic structure" for many categories like $\mathrm{G}roup$, $\mathrm{R}ing$, etc., is that these categories come equipped with an underlying set functor $\mathcal{C}\to\mathcal{S}et$ that has a left adjoint (the "free object" functor). Left adjoints respect colimits and right adjoints respect limits, which means that the underlying set of an inverse limit has to be the inverse limit of the underlying sets; so the construction is essentially forced as to what set it "should" be.

In general, if we let $\Delta\colon\mathbf{C}\to\mathbf{C}^{\mathbf{D}}$ be the diagonal functor that takes an object $X$ to the functor $\Delta(X)\colon\mathbf{D}\to\mathbf{C}$ that maps every object to $X$ and every arrow to $\mathrm{id}_X$; and takes each arrow $f\colon X\to Y$ in $\mathbf{C}$ to the natural transformation $\Delta(f)\colon\Delta(X)\to\Delta(Y)$ with value $f$ at all objects of $\mathbf{D}$, then a limit of a functor $F\colon\mathbf{D}\to\mathbf{C}$ is equivalent to an object $L$ that represents the contravariant functor $\mathbf{C}^{\mathbf{D}}(\Delta(-),F)\colon\mathbf{C}^{\rm op}\to\mathcal{S}et$. When every functor $\mathbf{D}\to \mathbf{C}$ has a limit, the $\lim\limits_{\leftarrow}$ construction is a right adjoint of the functor $\Delta\colon \mathbf{C}\to\mathbf{C}^{\mathbf{D}}$.

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Wow, thanks for the detailed answer! – ashpool Oct 18 '11 at 15:04

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