In Category in Context, the follow theorem is presented with the proof.
I think it's best to note that in the book, for small diagram $F: J \rightarrow Set$, $\lim F = Cone(1, F)$, where $Cone(1,F)$ is the set of cone with the trivial set as the summit.
I'm basically having hard time understanding the proof. As I don't think I understand any part of it, maybe working through an example and showing my confusion will help show what I'm thinking wrong.
So consider the pullback diagram:
We know explicitly that the limit of this diagram is the set $\{(a,b): f(a) = g(b) \}$
The theorem claims that this is a limit of equalizer diagram, where the domain is $\prod Fj = A \times B \times C$ and the codomain is $\prod F(cod f) = C \times C$
We also explicitly know the limit of the equalizer diagram, which is $\{(a,b,c): h((a,b,c)) = k((a,b,c)) \}$, for some $h$ and $k$.
Now, we must have unique isomorphism from $\{(a,b)| f(a) = g(b) \}$ and $\{(a,b,c): h((a,b,c)) = k((a,b,c)) \}$.
In this case, I'm guessing $h((a,b,c)) = (f(a), f(a))$ and $k((a,b,c)) = (g(b), g(b))$
But I'm having hard time relating this definition to the proof, so if anyone could help me, I would greatly appreciate it, thanks!
You've got a bit of messy notation here; for instance, you write $$\prod F(\mathrm{cod}(f))=C\times C$$ which puts $F(\mathrm{cod}(f))$ apparently living in the same category as $f$'s codomain. However $f$ lives in our diagram category $J$ and $F(\mathrm{cod}(f))$ lives in $\mathbf{Set}$. So I'm going to take $A,B,C,f,g$ as names for the objects of $J$, and just use $F(A),F(B),$ etc. for their images under the functor $F$.
So what's true is that $$\prod_{X\in\mathrm{Obj}J} F(X)=F(A)\times F(B)\times F(C)$$ and $$\prod_{j\in\mathrm{Mor}J}F(\mathrm{cod}(j))=F(\mathrm{cod}(id_A))\times F(\mathrm{cod}(id_B)) \times F(\mathrm{cod}(id_C))\times F(\mathrm{cod}(f))\times F(\mathrm{cod}(g)).$$ In the second product it's important to remember that we're indexing over all the morphisms, and that includes the identities. I continue to write it with the $\mathrm{cod}(-)$ notation for the objects because we will need to know which components of this product's 5-tuples correspond to which morphisms below.
You also don't quite have the right definition of the two morphisms we want to equalize; I don't think your text does a very good job at describing them, though it tries. Let $$\pi_Y:\prod_{X\in\mathrm{Obj}J} F(X)\to F(Y)$$ be the product projection for $Y\in\mathrm{Obj}J$. Then for each morphism $j$ of $J$ we have the projection $\pi_{\mathrm{cod}(j)}:\prod_{X\in\mathrm{Obj}J} F(X)\to F(\mathrm{cod}(j))$; t his induces a unique arrow $$c:\prod_{X\in\mathrm{Obj}J} F(X)\to \prod_{j\in\mathrm{Mor}J} F(\mathrm{cod}(j))$$ with action $(x,y,z)\mapsto (x,y,z,z,z)$.
There is also another cone given by the collection of morphisms $F(j)\circ\pi_{\mathrm{dom}(j)}$, which induces the map $$d:\prod_{X\in\mathrm{Obj}J} F(X)\to \prod_{j\in\mathrm{Mor}J} F(\mathrm{cod}(j))$$ with action $(x,y,z)\mapsto (x,y,z,F(f)(x),F(g)(y))$.
The limit then comes from the equalizer of $c,d$. Notice that, as a subset of $\prod_{X\in\mathrm{Obj}J} F(X)$, the equalizer is exactly the triples $(x,y,z)$ with $F(f)(x)=z=F(g)(y)$. We can then actually discard the $z$ component of the triples, since all we need to know is that $F(f)(x)=F(g)(y)$, and we get exactly the usual definition of the pullback.