A contravariant functor $\mathcal{F}:C\rightarrow Set$ is representable iff it has an universal object.


[$\Rightarrow$] As $\mathcal{F}$ is representable, then $\tau:h_X\rightarrow \mathcal{F}$ is an isomorphism for some $X\in C$.

Claim:$(X,\tau_X(1_X))$ is an universal object of $\mathcal{F}$ where $1_X\in h_XX$.

I was able to prove that claim and with that I prove the $[\Rightarrow]$ part, I am having problems with the reverse $[\Leftarrow]$. Can anybody give me some suggestion? I have done this:

$[\Leftarrow]$ Suppose $(X,a)$ with $X\in C$ and $a\in X$ is an universal object of $\mathcal{F}$, I need to show that it is representable. Lets consider the natural transformation $\tau:h_X\rightarrow \mathcal{F}$ for some $X\in C$, I have to show now that $\forall U\in C$ we have that $\tau_U:h_XU\rightarrow \mathcal{F}(U)$ is an isomorphism (or a bijection because I am on the set category), for this final statement can anybody give me any suggestion?

Thank you.


Suppose $(X, a) $ is a universal object for $\def\m{\mathcal} \m F:\m C\to\m Set$.
Then for every $(C, c) $ with $c\in\m FC$, there is a unique $f:X\to C$ with $\m Ff(a) =c$.

The representative object in this case will be $X$, and we can define $\tau:h_X\to \m F$ by $h_XC\ni f\mapsto \m Ff(a) \in\m FC$.
The uniqueness criterium of the universal property can either provide an inverse, or prove that it's a bijection componentwise, i.e. a natural isomorphism.

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