Category Theory: Inverse Limit in Sets

[Mandelbroth]

I think this looks like a homework problem, so I'll just put it here.

Homework Statement

Demonstrate that, for any index category $\mathscr{J}$ and any diagram $\mathcal{F}:\mathscr{J}\to\mathbf{Sets}$,

$$\varprojlim_{\mathscr{J}}A_j=\left\{a\in \prod_{j\in \operatorname{obj}( \mathscr{J})}A_j~\vert~(i,k\in\operatorname{obj}(\mathscr{J}),~a_i\in A_i,~a_k\in A_k, \text{ and } \phi_{ik}\in \hom_{\mathscr{J}}(i,k))~\implies~a_k=\mathcal{F}(\phi_{ik})(a_i) \right\}=A$$
along with the obvious projections, which I'll denote $l_i: A\to A_i$.

Homework Equations

I don't know how to make diagrams in TeX, so I'll just link to the universal property.

The Attempt at a Solution

Suppose $W$ has morphisms $w_i: W\to A_i$ that satisfy $w_k=\mathcal{F}(\phi_{ik})\circ w_i$. We wish to show the existence of a unique morphism $w: W\to A$ such that $w_i=l_i\circ w$.

My thought is that both $W$ and $A$ clearly map into the product $\prod A_j$. We even have the unique map from $A$ to $\prod A_j$, set inclusion, satisfying the universal property for the product. However, I don't know how to proceed. Any nudges in the right direction would be helpful. Thank you!

 

[micromass]

The maps $w_i:W\rightarrow A_i$ induce a unique map $w:W\rightarrow \prod_{i\in I} A$ such that $l_i\circ w = w_i$. This is essentially by definition of the product in the category.

Now show that $w$ actually maps into $A$, that is, that $w(W)\subseteq A$ and that it satisfies the universal property.

 

[Mandelbroth]

The maps $w_i:W\rightarrow A_i$ induce a unique map $w:W\rightarrow \prod_{i\in I} A$ such that $l_i\circ w = w_i$. This is essentially by definition of the product in the category.

Now show that $w$ actually maps into $A$, that is, that $w(W)\subseteq A$ and that it satisfies the universal property.

Let me see if I understand what you're saying. Let $\pi_i: \prod A_j\to A_i$ be the natural projection maps from the universal property defining the product. Then, $\pi_i\circ w=w_i$. But, since the $w_i$ commute with the induced maps $\mathcal{F}(\phi_{ik})$, and $A$ is defined as precisely the subset of $\prod A_j$ that does this, $w(W)\subseteq A$. Let $\rho: \prod A_j\to A$ be a left inverse of the inclusion map $i:A\to\prod A_j$, and let $w'=\rho\circ w:W\to A$. This map $w'$ is unique because $w(W)\subseteq A$, so the left inverse would take any element of $w(W)$ to its corresponding element of $A$.

Is this correct?

 

[micromass]

Yes, seems right!