Let the functor $F$ be $F(S)=S \times S$ (cartesian square).
For $A=\{a,b\}$ show $(a,b) \in A\times A$ is a universal element for the functor $F$.
Similarly $T=\{1,2\}$ show $(2,1) \in T \times T$ is a universal element for this functor $F$.
Show bijection between $A$ and $T$ given by the uniqueness of universal elements.
I am pretty new to this concept, please explain me in details so that I can fully understand. Thanks in advance.
For a functor $F:\mathcal C\to\mathbf{Set}$ a universal element is by definition a pair $\langle r,e\rangle$ where $r$ is an object in $\mathcal C$ and $e\in Fr$ such that for every pair $\langle d,x\rangle$ where $d$ again is an object of $\mathcal C$ and $x\in Fd$ there is a unique arrow $f:r\to d$ in $\mathcal C$ with $(Ff)(e)=x$.
Let us now show that universal elements are unique up to isomorphisms.
For this let it be that $\langle r,e\rangle$ and $\langle r',e'\rangle$ are both universal elements of functor $F$.
Then by definition there are (unique) arrows $f:r\to r'$ and $g:r'\to r$ such that $(Ff)(e)=e'$ and $(Fg)(e')=e$.
From this we find that $g\circ f:r\to r$ with $(F(g\circ f))e=(Fg)((Ff)e)=(Fg)e'=e$.
But we also have $1_r:r\to r$ with $(F1_r)e=e$ so the uniqueness allows us to conclude that $g\circ f=1_r$.
Likewise we find that $f\circ g=1_{r'}$ and proved is now that $f$ and $g$ are invertible arrows and are eachothers inverses.
In special case $\mathcal C=\mathbf{Set}$ this means exactly that $f$ and $g$ are bijections and are eachothers inverses.