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This might be a dumb question. If $C$ is an ordinary category, then for any $c \in C$ the covariant representable functor $\text{Hom}(c, -) : C \to \text{Set}$ preserves limits. However, it can happen that $c$ can be equipped with extra structure which in turn gives the morphisms out of $c$ extra structure, so that there is a "representable functor" $\text{Hom}(c, -) : C \to D$ where $D$ is a category equipped with a forgetful functor $F : D \to \text{Set}$ such that composing with the above gives the original representable functor.

In this situation, when does the functor into $D$ still preserve limits? How is this situation formalized? (Assume that $C$ is not enriched over $D$ in any obvious way.)

There are several examples of this coming from algebra, but the one that got me curious is the following. Let $C$ denote the homotopy category of pointed (path-connected?) topological spaces and let $S^1$ denote the circle with a distinguished point. I believe I am correct in saying that if the fundamental group functor $\pi_1 : C \to \text{Grp}$ is composed with the forgetful functor $U : \text{Grp} \to \text{Set}$, then $S^1$ represents the resulting functor $U(\pi_1(-))$. (The extra structure on $S^1$ that makes this possible is, if I'm not mistaken, a cogroup structure internal to $C$.) Can I conclude that $\pi_1$ preserves limits?

Edit: I've been told that the above example is problematic, so here's a simpler one. Let $C = \text{Set}$ and suppose that $c \in C$ is equipped with a morphism $f : c \to c$. Then by precomposition $\text{Hom}(c, d)$ is also equipped with such a morphism, so $\text{Hom}(c, -)$ has values in the category of dynamical systems. Does it preserve limits? Another example is my attempted answer to question #23188.

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If the forgetful functor creates limits, then I think you get what you want. In detail: Let $J$ be an index category, and let $J\stackrel{T}{\to}C\stackrel{F}{\to}D\stackrel{U}{\to}\mathbf{Set}$ be functors. Suppose that $UF$ preserves limits and $U$ creates limits. Suppose that $\tau\colon \ell\to T$ is a limiting cone in $C$. Since $UF$ preserves limits, $UF\tau\colon UF\ell\to UFT$ is a limiting cone. As $U$ creates limits, there is a unique lifting of $UF\tau$ to a cone in $D$, and this cone is a limiting cone. (to be continued...) – user2734 Apr 30 '10 at 23:07
(...cont'd) But $F\tau$ is such a lift, and hence we're done. Now, if I am not mistaken, for any $\tau$-algebra the forgetful functor to $\mathbf{Set}$ creates limits. (I once proved it for solving some exercise, but I'm not sure if my proof is correct, since it has been quite a while since I did it). – user2734 Apr 30 '10 at 23:07
The problem with your example is that the homotopy category doesn't even have limits. And even if you try to lift this into a model-category setting and use the homotopy limit, $\pi_1$ doesn't send those to limits (take a pullback of a fibration as an example, where you get a Mayer-Vietoris-like sequence involving all $\pi_i$). – Tilman May 1 '10 at 8:17
@Tilman: My comments referred only to the first part of the question. Is there something wrong with what I said? – user2734 May 1 '10 at 10:10
@unknown (google): no you're right. – Martin Brandenburg May 1 '10 at 10:48
up vote 5 down vote accepted

[Collecting my sporadic comments into one (hopefully) coherent answer.]

A more general question is as follows: For functors $C\stackrel{F}{\to}D\stackrel{U}{\to}E$ and for an index category $J$ such that $UF$ preserves $J$-limits, when does $F$ preserve $J$ limits?

A useful sufficient condition is that if $U$ creates $J$-limits, then in the above situation $F$ preserves $J$-limits. Proof: Let $T\colon J\to C$ be a functor, and suppose that $\tau\colon \ell\stackrel{\cdot}{\to} T$ is a limiting cone in $C$. Since $UF$ preserves $J$-limits, $UF\tau\colon UF\ell\stackrel{\cdot}{\to} UFT$ is a limiting cone in $E$. As $U$ creates $J$-limits, there is a unique lifting of $UF\tau$ to a cone in $D$, and this cone is a limiting cone. But $F\tau\colon F\ell\stackrel{\cdot}{\to} FT$ is such a lift, and hence we're done.

This condition is quite useful, because many forgetful functors are monadic, and monadic functors create all limits (by combining their definition on pp. 143--144 of Mac Lane and Ex. 6.2.2 on p. 142 of Mac Lane, or by Proposition 4.4.1 on p. 178 of Mac Lane--Moerdijk, or really by a comment of Tom Leinster from which I learned this :)).

For example, consider the category of all small algebraic systems of some type. From the AFT, we know that the forgetful functor to $\mathbf{Set}$ has a left adjoint, and it is the content of Theorem 6.8.1, p. 156 of Mac Lane that this forgetful functor is monadic.

Returning to the original question, this means that whenever the category $D$ is one of $\mathbf{Grp}$, $\mathbf{Rng}$, $\mathbf{Ab}$,... and $U\colon D\to \mathbf{Set}$ is the forgetful functor, then for any $J$, $UF$ preserves $J$-limits implies $F$ preserves $J$ limits. In particular, if $UF$ is a representable functor (and hence preserves all limits), then $F$ preserves all limits.

Next, let me try to comment on your motivating examples (the one from Q. 23188 and the one from the 'Edit' part of the current question.)

Regarding your example in Q. 23188: Unfortunately I know nothing of Hopf algebras, so I can't understand all the details of your construction. If I understand correctly, you construct a functor $F\colon\mathbf{Rng}\to\mathbf{Grp}$ whose composition with the forgetful functor $U\colon \mathbf{Grp}\to \mathbf{Set}$ is representable. If this is indeed the case, then by the above $F$ itself preserves all limits.

[EDIT: corrected the part concerning the last example.]

Finally, regarding your example in the edited question: While I know nothing of dynamical systems, from a quick glance at Terence Tao's blog it seems that the category of dynamical systems is the category whose objects are pairs $\langle X,f\colon X\to X\rangle$ with $X$ a (small) set and whose arrows $\phi\colon\langle X, f\rangle\to\langle Y, g\rangle$ are those functions $\phi\colon X\to Y$ with $g\circ\phi =\phi\circ f$.

To show that the above sufficient condition works in this case, we would like to show that the forgetful functor to $\mathbf{Set}$ crates limits. More generally, we will show that if $C$ is a category and $D$ is the category whose objects are pairs $\langle x,f\colon x\to x\rangle$ (where $x\in\operatorname{obj}(C)$, $f\in\operatorname{arr}(C)$), and whose arrows $\phi\colon \langle x,f\rangle\to \langle y,g\rangle$ are those arrows $\phi\colon x\to y$ with $g\circ\phi =\phi\circ f$, then the forgetful functor $U\colon D\to C$ creates limits.

[I'm sure that this follows from some well-known result, but since I don't see it, I'll just continue with a direct proof.]

So, let $J$ be an index category, let $F\colon J\to D$ be a functor, and suppose that $\tau\colon x\stackrel{.}{\to} UF$ is a limiting cone in $C$. We would like to show that there exists a
unique cone $\sigma\colon L\stackrel{.}{\to} F$ in $D$ such that $U\sigma=\tau$, and that this unique cone is a limiting cone.

For uniqueness, suppose that $\sigma\colon L\stackrel{.}{\to} F$ satisfies $U\sigma = \tau$. Write $F_j:=\langle y_j,f_j\rangle$. Then we must have for all $j$ $$ \sigma_j=(x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j) $$ for some $f\colon x\to x$ (hence we immediately see that $\sigma$ is determined up to $f$). Now, since by the above we see that $\tau_j$ must be an arrow $$ (x\stackrel{f}{\to}x)\stackrel{\tau_j}{\to}(y_j\stackrel{f_j}{\to}y_j) $$ of $D$, the following diagram must be commutative for all $j$: $$ \begin{matrix} x & \stackrel{\tau_j}{\longrightarrow} & y_j =UF_j\\ f\downarrow & & f_j\downarrow\\ x&\stackrel{\tau_j}{\longrightarrow} & y_j = UF_j. \end{matrix} \quad \text{(Diagram 1)} $$

Now we claim that the $\to\downarrow$ part of the above diagram forms a cone to $UF$, that is, we claim that the family $\{f_j\tau_j\}$ forms a cone $x\stackrel{.}{\to} UF$. Indeed, for an arrow $g:j\to j'$ of $J$, consider the following diagram: $$ \begin{matrix} &&&&x\\ &&&\stackrel{\tau_j}{\swarrow}&&\stackrel{\tau_{j'}}{\searrow}\\ &&y_j && \stackrel{UFg}{\longrightarrow} && y_{j'}\\ &\stackrel{f_j}{\swarrow} &&&&&&\stackrel{f_{j'}}{\searrow}\\ y_j&&&&\stackrel{UFg}{\longrightarrow}&&&&y_{j'} \end{matrix} $$

The upper triangle is commutative because $\tau$ is a cone to the base $UF$, and the lower trapezoid is commutative because $F$ is a functor, and hence $Fg$ is an arrow $F_j\to F_{j'}$ in $D$. Hence the outer triangle commutes, as required. From the universality of $\tau$, it follows that there is a unique $f$ for which Diagram 1 is commutative, and we have uniqueness.

For existence, we can take $f$ to be the unique arrow $x\to x$ for which Diagram 1 is commutative, and we get a cone $$ \sigma=\{\sigma_j=\tau_j\colon (x\stackrel{f}{\to}x)\to F_j=(y_j\stackrel{f_j}{\to}y_j)\} $$ with $U\sigma=\tau$. We claim that this is a limiting cone.

To see this, let $\alpha\colon(z\stackrel{g}{\to}z)\stackrel{.}{\to}F$ be a cone, so that for all $j$ the following diagram is commutative: $$ \begin{matrix} z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\ g\downarrow & & f_j\downarrow\\ z &\stackrel{\alpha_j}{\longrightarrow} & y_j. \end{matrix} \quad\text{(Diagram 2)} $$

Then $U\alpha$ is a cone $z\stackrel{.}{\to} UF$ in $C$, and by the universality of $\tau$ there exists a unique arrow $h\colon z\to x$ for which the following diagram is commutative for all $j$: $$ \begin{matrix} z & \stackrel{\alpha_j}{\longrightarrow} & y_j\\ h\downarrow& \stackrel{\tau_j}{\nearrow}\\ x& \end{matrix}\quad\text{(Diagram 3)} $$

If this $h$ is an arrow $(z\stackrel{g}{\to}z)\to (x\stackrel{f}{\to}x)$ in $D$, then we're done. In other words, all that remains to do is to show that the outer rectangle of the following diagram is commutative: $$ \begin{matrix} z && \stackrel{h}{\longrightarrow} && x\\ & \stackrel{\alpha_j}{\searrow} && \stackrel{\tau_j}{\swarrow}\\ && y_j\\ g\downarrow&& \downarrow f_j && \downarrow f\\ && y_j\\ & \stackrel{\alpha_j}{\nearrow} && \stackrel{\tau_j}{\nwarrow}\\ z && \stackrel{h}{\longrightarrow} && x\\ \end{matrix} $$ Now, the left trapezoid is just Diagram 2, the upper and lower triangles are just Diagram 3, and the right trapezoid is commutative for all $j$ by the definition of $f$. It follows that both paths of the outer rectangle have the same composition with the limiting cone $\tau$, and hence the outer rectangle is commutative, as required.

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Oops, the part concerning dynamical systems is wrong: the category of algebras for the identity monad on $\mathbf{Set}$ is just $\mathbf{Set}$. But I think that I can prove creation directly in this example, and I will soon either send the correction or declare the last example as open. – user2734 May 10 '10 at 9:01
OK, now I have a (hopefully) correct direct proof. – user2734 May 10 '10 at 20:11

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