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I've seen the following theorem around in various forms:

To give an object $A \in \mathcal{C}$ the structure of a $\Omega$-algebra object in $\mathcal{C}$ is equivalent to giving a lift of the contravariant hom-functor $\mathcal{C}(-,A) \colon \mathcal{C} \to Set$ to a functor $\mathcal{C} \to \Omega$.

Here $\Omega$ denotes the category of $\Omega$-algebra objects in $Set$.

However, despite much searching I've been unable to find a proof in the reverse direction. My particular question is given a lift of the hom-functor, how can we define the maps giving $A$ the structure of a $\Omega$-algebra object.

EDIT: After the answer by Dimitri, here's my attempt at a proof:

Let $h_A = \mathcal{C}(-,A) \colon \mathcal{C}\to Set$ and suppose that this lifts.

$h_A$ is an $\Omega$-algebra object in the functor category $[\mathcal{C}, Set]$. Let's suppose for an example that our $\Omega$-algebra, has a multiplication $\mu$. Then we have a natural transformation $\mu \colon h_A \times h_A \to h_A$.

The Yoneda embedding now tells us we have an isomorphism $Nat(h_{A \times A}, h_A) = \mathcal{C}(A \times A, A)$. Since $h_{A \times A}$ is isomorphic to $h_A \times h_A$ we see our natural transformation specifies a map $A \times A \to A$.

Thanks

Will

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Giving a functor $C \rightarrow \Omega$ is the same as giving an object with an $\Omega$-algebra structure in the category of functors $[C, Set]$. Having observed this the theorem follows form Yoneda and the fact that it preserves products.

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  • $\begingroup$ Thanks Dimitri, I've been trying to prove this myself, as you can see in the edit, but have been running into a little trouble, could you point me in the right direction? Ideally, I'd like a constructive proof for a concrete setting, e.g. for $\Omega$ being algebras over a commutative ring. Is there a more straightforward way to do this then trying to unwind everything the Yoneda lemma tells us? $\endgroup$
    – WMycroft
    Commented Aug 14, 2014 at 18:36
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    $\begingroup$ $h_A\times h_A$ is isomorphic to $h_{A\times A}$. This is because Yoneda preserves products, and pretty much the definition of a product. Then you use Yoneda $Nat(h_{A\times A}, h_A) \cong C(A\times A, A)$. That is the fact that Yoneda is fully faithful $\endgroup$
    – Dimitri Chikhladze
    Commented Aug 14, 2014 at 18:49
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    $\begingroup$ Btw, perhaps this question was more appropriate for math.stackexchange $\endgroup$
    – Dimitri Chikhladze
    Commented Aug 14, 2014 at 18:53
  • $\begingroup$ Got it, thanks a lot for your help. Do you have any insight over how easy it would be to do this in a constructive manner? The other direction in the equivalence is very easy to do constructively. I didn't realise math.stackexchange and mathoverflow were different things, and agree with you. Is it possible to move this question? $\endgroup$
    – WMycroft
    Commented Aug 14, 2014 at 19:49
  • $\begingroup$ You're wellcomed :) Not sure what do you mean by constructive, but e.g. to get the operation $A\times A \rightarrow A$ you evaluate $h_A\times h_A(A\times A) \rightarrow h_A (A\times A)$ at $(p_1, p_2)$, where $p_1$ and $p_2$ are the two projections $A\times A \rightarrow A$. $\endgroup$
    – Dimitri Chikhladze
    Commented Aug 14, 2014 at 20:10

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