How does one show that $A \cong T \times A $ when $T$ is terminal?

Question

I want to show:

$$ A \cong T \times A $$

so for that I wanted to do $f;g = 1_A$ and $g;f = 1_{T \times A}$. The first one is easy $f;g = 1_A$ because we know $T \times A$ is a limit so for any other cone $(C,\{\gamma_A,\gamma_T\})$ of the disconnected objects $T$,$A$ we have the factorization:

$$ \gamma_A = f;g$$

in particular for $\gamma_A = 1_A$. So it works.

But how do we show the other direction? I assume something to do with the fact that $T$ is terminal (i.e. admits unique morphisms to it from any object) but this simple fact is escaping me right now. Any idea?

Answer 1

Here's the brute-force approach I mentioned in the comments. This is definitely not the slickest way.

Let $\pi_i : A_1\times A_2\to A_i$ be projections of the product $A_1\times A_2$. Let $\langle f,g\rangle : B\to A_1\times A_2$ where $f: B\to A_1$ and $g : B\to A_2$ be the unique morphism (for any $B$, $f$, and $g$) whose existence is guaranteed by the universal property. The universal property can then be stated as: $\pi_i\circ\langle f_1,f_2\rangle = f_i$ and $\langle \pi_1\circ g,\pi_2\circ g\rangle = g$ for any pair of arrows $f_i: B\to A_i$ for any $B$ and any $g : C \to A_1\times A_2$ and any $C$.

Let $1$ be a terminal object and $!_A : A\to 1$ be the unique morphism (for each $A$) whose existence is guaranteed by the universal property of terminal objects. The universal property can then be stated as: $!_A = f$ for any $f : A\to 1$ for any $A$.

Claim: $\pi_2 : 1\times A \to A$ and $\langle !_A,id_A\rangle : A\to 1\times A$ form an isomorphism, i.e. are mutually inverse. We need to show $$\begin{align}\pi_2\circ\langle!_A,id_A\rangle & = id_A\qquad \text{ and }\\ \langle !_A,id_A\rangle\circ\pi_2 & = id_{1\times A}\end{align}$$ The first holds immediately via the universal property for products. For the second, we use the second part of the universal property of products which implies that if $\pi_i\circ f=\pi_i\circ g$ for $i\in\{1,2\}$ then $f=g$. (Why?) Let's try to prove $$\begin{align}\pi_1\circ\langle !_A,id_A\rangle\circ\pi_2 & = \pi_1\circ id_{1\times A}\\ \pi_2\circ\langle !_A,id_A\rangle\circ\pi_2 & = \pi_2\circ id_{1\times A}\end{align}$$ The second immediately follows from the universal property for products. For the former we get $$\begin{align}\pi_1\circ\langle !_A,id_A\rangle\circ\pi_2 & = {!_A}\circ \pi_2 \\ & = {!_{1\times A}} \\ & = \pi_1 \\ & = \pi_1\circ id_{1\times A}\\\end{align}$$ The $!_A\circ\pi_2 = {!_{1\times A}}$ and $!_{1\times A} = \pi_1$ both used the universal property of terminal objects. As mentioned, we finish by using the universal property of $1\times A$ to get $$\begin{align}\langle\pi_1\circ\langle!_A,id_A\rangle\circ\pi_2,\pi_2\circ\langle!_A,id_A\rangle\circ\pi_2\rangle & =\langle\pi_1\circ id_{1\times A},\pi_2\circ id_{1\times A}\rangle\\ & = id_{1\times A}\end{align}$$

Answer 2

The arrows from some object $B$ to $T\times A$ correspond naturally to pairs of arrows $(f,g)$ where $f:B\to T$ and $g:B\to A$. But there's only one arrow from $B$ to $T$, so the arrows $B\to T\times A$ correspond naturally to the arrows from $B\to A$. So there's a natural equivalence of Hom-functors $\text{Hom}(-,A)$ and $\text{Hom}(-,T\times A)$. So $A$ is isomorphic to $T\times A$ (basically by Yoneda).

Answer 3

For a proof from first principles (with functors and Yoneda officially not yet in sight), have a look at Theorem 26 and then Theorem 27(i) of this Gentle Introduction.

Answer 4

Show that $A$ satisfies the universal property for $T\times A$. Let $\pi_T=!$ (the unique map to $T$), and let $\pi_A=1_A$ (the identity on $A$). Now let $f:B\to T$ and $g:B\to A$ be any arrows. Define $h:B\to A$ by $h=g$. Then $\pi_T\circ h=f$ since there is only one map $B\to T$ and $\pi_A\circ h=1_A\circ g=g$. Furthermore, if $h':B\to A$ is another such map satisfying these relations, then $$h'=1_A\circ h'=\pi_A\circ h'=\pi_A\circ h=1_A\circ h=h,$$ which gives uniqueness. Since products are unique up to isomorphism, we conclude $A\cong T\times A.$

Why does everyone think you need Yoneda to prove something so basic?