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EDIT: The original question had a trivial answer: it's just a coproduct. New question below New Question: As shown below, in the category of commutative unital rings, the coproduct of a ring $R$ with $\mathbb Z\left[x\right]$ (which represents the forgetful functor to sets) is the polynomial ring $R\left[x\right]$. In what other concrete categories does the coproduct of an object $A$ with the object $Z$ ($Z$ is the object that represents the forgetful functor) give a useful construction.

Reformulating the question: When does the free object on the point in the co-slice category of maps $A\to B$ give a useful construction? (from giuseppe's comment below this question)

Given a ring (commutative, unital) $R$, the polynomial ring $R\left[x\right]$ is the initial pointed $R$-algebra, in the sense that given any other $R$-algebra $R\to A$ and a point $a\in A$ we have a unique $R$-algebra homomorphism $p\colon R\left[x\right]\to A$ such that $p\left(x\right)=a$.

Since the forgetful functor $\mathrm{CRing}\to\mathrm{Sets}$ is $\mathrm{Hom_{CRing}}\left(\mathbb Z\left[x\right],-\right)$ we can equivalenty think of $R\left[x\right]$ to be initial ring that is both an $R$-algebra and a $\mathbb Z\left[x\right]$-algebra. In diagrammatic form, we could say $\require{AMScd}$ \begin{CD} R @>\iota>>R\left[x\right] @<x\mapsto x<<\mathbb Z\left[x\right]\\ @V \mathrm{id}_RVV @VV\exists! V @VV\mathrm{id}_{\mathbb Z\left[x\right]}V\\ R @>>\phi> A@<<a<\mathbb Z\left[x\right] \end{CD}

I was thinking of this definition in any concrete category. So if I have a category $\mathcal C$ with a faithful functor $U\colon\mathcal C\to\mathrm{Sets}$ which is representable, that is $U\simeq\mathrm{Hom}_\mathcal C\left(Z,-\right)$ for some object $Z$ in $\mathcal C$. Then for any object $A$, we define the polynomial object $P$ over $A$ to be universal in the same sense as above, i.e.,

\begin{CD} A @>\iota>>P_A @<f<<Z\\ @V \mathrm{id}_AVV @VV\exists! V @VV\mathrm{id}_ZV\\ A @>>\phi> Q@<<g<Z \end{CD} EDIT: This is obviously a coproduct. Ignore the questions below. I was wondering what this would correspond to in other concrete categories with representable forgetful functors. I also wanted to know some example of this construction in other categories. Are these 'polynomial objects' things that already have names in other specific categories?

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    $\begingroup$ I don't quite understand your diagram but wouldn't it just be the coproduct of Z with A (assuming co products exist). $\endgroup$
    – Asvin
    Commented Jul 16, 2020 at 20:15
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    $\begingroup$ Someone should write a CW answer and Chetan can accept it, to keep this Q off the 'unanswered' list. $\endgroup$
    – David Roberts ♦
    Commented Jul 16, 2020 at 23:20
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    $\begingroup$ @DavidRoberts I have edited the question and replaced it by a new question that came up. $\endgroup$
    – Chetan Vuppulury
    Commented Jul 16, 2020 at 23:36
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    $\begingroup$ If $U$ is representable then it has a left adjont $F,$ since the representing object of $U$ has the universal property of the free object on the point and then for every other set $A$ you build $F(A)$ as $\sqcup_{A}F(*)$ $\endgroup$
    – display llvll
    Commented Jul 17, 2020 at 14:05
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    $\begingroup$ I think your question is too broad to have an answer. You are just taking the composition of two left adjoints; you have the left adjoint to your functor $U$ and then you compose it with the left adjoint to the forgetful functor $(A \to B ) \mapsto B$, which is just given by taking the coproduct with $A$. So, a stupid answer would be that this is useful whenever the category of maps $A \to B$ is interesting. $\endgroup$
    – display llvll
    Commented Jul 18, 2020 at 3:31

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