Why having finite products and equalizers implies having terminal object

category-theory

So, I am reading Steve Awodey "Category Theory", and what he claims is that if category has finite products and equalizers then it has terminal object. I am not so sure why is this true. I mean, he says "a category", which can be large. So, a terminal object is a product of all objects in a category, so possibly infinite products also have to exist. Why is this true? Thanks!

Best Answer

The terminal object is not the product of all objects, it is the product of no objects. Already having finite products implies having a terminal object, by taking the product of the empty set of objects.

Why is the terminal object the empty product? See this question.

Related Solutions

Finite Products vs Finite Coproducts in Additive Categories

Note that a product $A \times B$ is the same as a pull-back diagram

A x B -> B
  |      |
  v      v
  A ---> 0

with maps $p:A \times B \to A$ and $q: A \times B \to B$. In particular there is a map $i: A \to A \times B$ such that $pi = 1_{A}$ and $qi = 0$ as well as a map $j: B \to A \times B$ such that $qj = 1_{B}$ and $pj = 0$. Now note that $p(ip + jq) = pip + 0 = p$ and $q(ip + jq) = q$ so that $ip + jq = 1_{A \times B}$.

Let us check that $i: A \to A \times B$ and $j: B \to A \times B$ define a coproduct. Given $f: A \to D$ and $g: B \to D$ we get a map $d: A \times B \to D$ by setting $d = fp + gq$. Since $di = (fp + gq)i = fpi +gqi = f$ and $dj = g$ it remains to prove uniqueness of $d$. But this is clear as any other such map will satisfy $(d-d')1_{A \times B} = (d-d')(ip+jq) = 0$. Summing up, we have proved that the product of two objects is also a coproduct.

Now if we want to say that finite products and coproducts exist and coincide, we need a zero object, since otherwise the empty product (the terminal object) and the empty coproduct (the initial object) would not coincide.

[Math] Products and pullbacks imply equalizers

Yes, products and pullbacks imply equalizers. The equalizer of $f,g: A \to B$ is the pullback of $(1,f)$ and $(1,g): A \to A\times B$. A cone $C$ over the pullback diagram of $(1,f)$ and $(1,g)$ has maps $\pi_1,\pi_2: C \to A$ and by commutativity of the diagram we know $\pi_1 = \pi_2$ and $f \circ \pi_1 = g \circ \pi_2$. This is clearly equivalent to a cone over the equalizer diagram of $f,g: A \to B$.

Best Answer

Related Solutions

Finite Products vs Finite Coproducts in Additive Categories

[Math] Products and pullbacks imply equalizers

Related Question