I'm fairly new to category theory, and currently working through Leinster's category theory problem sets. I'm stuck on the following:
Let $C$ a small category, $C_0$ the objects of $C.$
Describe a left adjoint $F$ to the forgetful functor $U:[\mathbb{C},Set] \rightarrow Set^{\mathbb{C}_0},$ where $Set^{\mathbb{C}_0}$ is the product of $\mathbb{C}_0$ copies of $Set.$
I'm guessing that $U$ is defined on objects by, for any $G: \mathbb{C} \rightarrow Set,$ $U(G)=\times_{c \in \mathbb{C}}G(c),$ and on transformations by, for any $\alpha:G \rightarrow H,$ $F(\alpha):=\times_{c \in \mathbb{C}}\alpha_c.$
Now let $Q \in Set^{\mathbb{C}_0},$ and $\varphi \in [\mathbb{C},Set].$
We want to define $F$ so that we have an isomorphism $Hom(F(Q),\varphi) \cong Hom(Q,U(\varphi)).$ Well, by construction each morphism $R$ in $Hom(Q,U(\varphi))$ can be thought of as $R=\times_{c \in C}R_c,$ where for each $c,$ $R_c:Q \rightarrow \varphi(c).$
My first thought was to have $F(Q)(c)=Q$ for all $c,$ and send all morphisms in $C$ to $id_Q.$ If we forget temporarily that $\alpha \in Hom(F(Q),\varphi)$ has to be a natural transformation, and just think of $\alpha$ as a collection $\{\alpha_c:F(Q)(c)=Q \rightarrow \varphi(c)|c \in C\},$ then we would have an obvious correspondance between the two morphism sets.
But then once you consider naturality of $\alpha,$ you would get $\varphi(f) \circ \alpha_c=\alpha_{c'}$ for all $c,c' \in \mathbb{C}$ and all $f:c \rightarrow c',$ which obviously won't work.
I feel like I'm making this way more difficult than it should be. Any ideas?
Edit:
Thanks to all for the helpful comments. I've been thinking about this problem on and off, and I still don't have a handle on it. As I understand it now, a morphism in $Hom(Q,U(\varphi))$ is simply a collection of functions $\{f_c:Q_c \rightarrow \varphi(c)\}_{c \in C}.$ I have no idea how to construct a functor $F$ such that the natural transformations from $F(Q) \rightarrow \varphi$ are even in one to one correspondence with the aforementioned collections of functions.
Once you have resolved the confusion over sequences of sets versus product sets, here's a hint as to how you might proceed with the problem...
Hint: let us think in terms of this heuristic (which I got from George Bergman) for constructing a left adjoint: suppose we fix $X \in Set^{C_0}$, and then we have some functor $G \in [C, Set]$ along with a morphism $\alpha : X \to U(G)$ in $Set^{C_0}$. This means that for each $c \in C_0$, we have a morphism $\alpha_c : X_c \to G(c)$. Then:
(a) "What objects must exist in $G$?"
(b) What relations must exist between these objects?
For the first question, we'll interpret that as $U(G)$ being an indexed family of sets (indexed by $C_0$), and determining what objects must exist in each set in this family. So, after some experimentation, you will find that there must be objects like $\alpha_c(x) \in G(c)$ for each $x \in X_c$; $G(f)[\alpha_c(x)] \in G(d)$ for $x \in X_c, f \in \operatorname{Hom}_C(c, d)$; $G(g)[G(f)[\alpha_c(x)]] \in G(e)$ for $x \in X_c, f \in \operatorname{Hom}_C(c, d), g \in \operatorname{Hom}_C(d, e)$; etc.
For the second question, relations will come from the fact that $G$ is a functor. So, for example, we will get relations like:
$$G(\operatorname{id}_c)(\alpha_c(x)) = \alpha_c(x), x \in X_c$$
from the identity condition, and
$$G(g)[G(f)[\alpha_c(x)]] = G(g \circ f)(\alpha_c(x)), x \in X_c, f \in \operatorname{Hom}_C(c, d), g \in \operatorname{Hom}_C(d, e)$$
from the composition condition.
Now, the rest of the heuristic goes: try to make an object in the appropriate category (in this case, try to make a functor) just from the necessary generators and relations that you have found. (It will often help to use the relations to reduce the number of generators as appropriate.) If you've found "enough" generators and relations, then you should be able to construct the category's necessary "operations" (in this case, the part of the functor acting on morphisms) and prove it satisfies the "identities" of the category (in this case, the identity and composition conditions); if not, the failures could either give some useful hints as to what generators or relations you might be "missing", or else some hints as to why the functor might not have an adjoint.
Usually, once you have finished this step, proving that you have constructed an adjoint functor will be straightforward.
(Note that the heuristic described above applies most directly when the source category of the functor, a.k.a. the target category of the desired adjoint functor, is a variety of algebras, for example, the category of groups. For other categories, you might have to stretch the interpretation as I did here, but depending on the category involved, it can still be a useful point of view as it was here.)