Vrftsjtryk

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?

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

Let $pi_i : A_1times A_2to A_i$ be projections of the product $A_1times A_2$. Let $langle f,grangle : Bto A_1times A_2$ where $f: Bto A_1$ and $g : Bto A_2$ 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: $pi_icirclangle f_1,f_2rangle = f_i$ and $langle pi_1circ g,pi_2circ grangle = g$ for any pair of arrows $f_i: Bto A_i$ for any $B$ and any $g : C to A_1times A_2$ and any $C$.

Let $1$ be a terminal object and $!_A : Ato 1$ 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 $f : Ato 1$ for any $A$.

Claim: $pi_2 : 1times A to A$ and $langle !_A,id_Arangle : Ato 1times A$ form an isomorphism, i.e. are mutually inverse. We need to show $$begin{align}pi_2circlangle!_A,id_Arangle & = id_Aqquad text{ and }\
langle !_A,id_Aranglecircpi_2 & = id_{1times A}end{align}$$
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 $pi_icirc f=pi_icirc g$ for $iin{1,2}$ then $f=g$. (Why?) Let's try to prove $$begin{align}pi_1circlangle !_A,id_Aranglecircpi_2 & = pi_1circ id_{1times A}\ pi_2circlangle !_A,id_Aranglecircpi_2 & = pi_2circ id_{1times A}end{align}$$ The second immediately follows from the universal property for products. For the former we get $$begin{align}pi_1circlangle !_A,id_Aranglecircpi_2 & = {!_A}circ pi_2 \ & = {!_{1times A}} \ & = pi_1 \ & = pi_1circ id_{1times A}\end{align}$$
The $!_Acircpi_2 = {!_{1times A}}$ and $!_{1times A} = pi_1$ both used the universal property of terminal objects. As mentioned, we finish by using the universal property of $1times A$ to get $$begin{align}langlepi_1circlangle!_A,id_Aranglecircpi_2,pi_2circlangle!_A,id_Aranglecircpi_2rangle & =langlepi_1circ id_{1times A},pi_2circ id_{1times A}rangle\ & = id_{1times A}end{align}$$

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

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

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

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.