# Forgetful Functors Create Limits

I'm working on the following problem but I can't seem to make any headway.

A widget is a set $A$ with elements $0,1 \in A$, a ternary operation $[-,-,-]: A^3 \to A$, and for each rational number $\lambda$, a unary operation $\lambda\cdot - : A \to A$ satsifying the axioms

1. $[a,0,b]=[3b/4,a/4,1]$

2. $[[a,b,a],c,a] = 0$

3. $\lambda\cdot (\mu\cdot a) = (\lambda\mu)\cdot a$

4. $(\lambda+\mu)\cdot a = [\lambda\cdot a,\mu \cdot a, (5-\mu)\cdot a]$

for all $a,b,c \in A$ and rational $\lambda,\mu$. If Widget is the category of widgets and their homomorphisms. Then show the forgetful functor $U: \mathbf{Widget} \to \mathbf{Set}$ creates limits.

I realize the algebraic structure of widgets in this case are incidental and such a proof should be equally applicable to the category of rings, groups, modules, etc. However given a limit $(L, \lambda_j: L \rightarrow UD(j))$ where $D: \mathcal{J} \to \mathbf{Widget}$ is a diagram. I don't know how to find a widget (or group or ring) M such that $U(M) \cong L$ nor do I know how to find pre-images of the legs of the cone with summit $L$ under the functor U, such that M forms a cone in $\mathbf{Widget}$.

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It might help to look at an explicit construction of limits in $\mathbf{Set}$. (Or just products to start off with.) –  Colin McQuillan Mar 13 '13 at 11:35
If it creates limits, then the underlying set of $M$ is $L$. So all you have to do is put the appropriate algebraic structure on $L$. (If widgets are too bizarre for your taste, try doing it for groups first.) –  Zhen Lin Mar 13 '13 at 14:06
So from what I can gather from these hints, a limit $L$ is of the form $L = \{{x}_{j\in J}\in \prod_{j\in J}UD(j): D(\alpha)(x_j) = x_{j'} \forall \alpha \in \mathrm{mor}{\mathcal{J}}\}$ where $J = \mathrm{ob}{\mathcal{J}}$. Now since a widget is just a set with some added structure, then I can construct a limit in the category of widgets in an identical fashion using the preimage of the sequences $\{x_j\}$. I have an additional question: If I have a forgetful functor between two "algebraic" categories $U: \mathcal{C}\to \mathcal{D}$. Does the same argument show this functor creates limits? –  Richard Leyland Mar 13 '13 at 14:32
Yes, you could do it like that if you want. Alternatively, note that you have a commutative triangle of forgetful functors, all of which preserve limits, and two of them create limits. –  Zhen Lin Mar 13 '13 at 18:33
Good problem, could you please give a reference where does it come from? –  magma Mar 14 '13 at 10:55
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Let $\mathcal A$ be the category of some given (equationally defined) algebraic structures and their homomorphisms. Then consider the bigger category $\mathcal U$ connecting $\mathcal{Set}$ to $\mathcal A$ via the underlying functor $U:\mathcal A\to\mathcal{Set}$, that is, $\mathcal U$ contains the disjoint union of $\mathcal{Set}$ and $\mathcal A$ as full subcategories and (so called) $S\dashrightarrow A$ heteromorphisms from objects $S$ in $\mathcal{Set}$ to objects $A$ in $\mathcal A$, as functions $S\to UA$. (This is basically a view of the profunctor $U^*:\mathcal{Set}^{op}\times\mathcal A\to\mathcal{Set}:\ \langle S,A\rangle\mapsto \mathcal{Set}(S,UA)$.)
Now, by definition, $\mathcal{Set}$ is a coreflective subcategory of $\mathcal U$, meaning that each $A$ has a coreflection arrow $UA\dashrightarrow A$ (namely $1_{UA}$). By construction of free algebras, we can see that, in addition, $\mathcal A$ is a reflective subcategory of $\mathcal U$ (the reflection is given by the free algebra functor, the left adjoint of $U$). Using this latter fact one can easily show that limits in $\mathcal A$ are actually limits in $\mathcal U$.
Finally, let $\mathcal D$ be a diagram in $\mathcal A$ with limit object $L$, and let $S\leadsto U\mathcal D$ be a cone over $U\mathcal D$ in $\mathcal{Set}$. Connect $UX\dashrightarrow X$ by the coreflection arrows for each object of $X$ of $\mathcal D$. Then, as $L$ is the limit of $\mathcal D$ within $\mathcal U$, we have a unique $S\to L$ by the limit property, and this leads to a unique $S\to UL$ by the coreflection property.