I need to show the following conditions are equivalent for a category $ C $
(a) $ C $ has binary products, equalizers, and a terminal object;
(b) $ C $ has pullbacks and a terminal object;
(c) $ C $ has finite limits
I know that if a category has pullbacks and a terminal object then it has binary products. I also know that if a category has pullbacks then it has equalizers.
I know that a category has all finite limits iff it has finite products and equalizers. IE if it has pullbacks and a terminal object.
I need to show one implies the others, right?
here is what I have:
If $C$ has pullbacks and terminal object 1, then to show $C$ has binary products, consider the pullback of $A, B \in C$
$$ \begin{array}{ccc}
A \times B & \pi _{2} \rightarrow & B\\
\pi_{1} \downarrow & & \downarrow t_B\\
A & t_A \rightarrow & T
\end{array} $$
$ \pi _{1} , \pi _{2} $ is a product cone over $A$ and $B.$ This is because for $p_{1}: P \rightarrow A$ and $p_{2}: P \rightarrow B$ we have $t_{A} \circ p_{1} = t_{B} \circ p_{2}$ where $t$ is the terminal object
therefore we have unique $h: P \rightarrow A \times B$ with $\pi _{1} \circ h = p_{1} ; \pi_{2} \circ h = p_{2}$
I know having pullbacks means you have equalizers, I had to prove that in a previous question.
So I have shown that (b) implies (a).
All I need is to now show (a) implies (c) right?
any
Best Answer
Given binary products, equalizers, and a terminal object, first use induction and binary products to show that you have products for any finite number $n\geq1$ of objects. You also have the product of $0$ objects, namely the terminal object, so you have all finite products. In particular, given a finite diagram, say with $V$ as its set of vertices (objects of your category), with $E$ as its set of edges (morphisms of your category), and with each edge $e$ having source $s(e)$ and target $t(e)$ in $V$, you can form the product $X$ of all the objects in $V$ and the product $Y$, indexed by $E$, of the factors $t(e)$. Consider the two morphisms $X\to Y$ defined as follows by giving their components at each edge $e$. The first has, as its $e$-component, the morphism $X\to t(e)$ that is simply the $t(e)$-th projection of the product $X$. The second has as its $e$-th component the composite of the $s(e)$-th projection of $X$ with the morphism $e:s(e)\to t(e)$. Let $L\to X$ be the equalizer of these two morphisms. Composing it with the projections $X\to v$ for all vertices $v$ of your diagram, you get morphisms $L\to v$ for all $v\in V$. It is straightforward to check that these morphisms constitute a limit cone for your diagram.