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I need an help in the proof of the following fact:

the category $Mod_{\mathcal{T}}$ is reflective in the category $Fun(\mathcal{T},Set)$

Here $\mathcal{T}$ is an algebraic theory, i.e. a category with a denumerable set of objects $\{T^0, T^1, \ldots, T^n,\ldots\}$, each $T^n$ being the $n$-th product of $T^1$. A $\mathcal{T}$-model is any functor $G:\mathcal{T}\longrightarrow Set$, which is finite product-preserving. The category $Mod_{\mathcal{T}}$ is the full subcategory of $Fun(\mathcal{T},Set)$, whose objects are all the finite product-preserving functors.

Being reflective in the functor category $Fun(\mathcal{T},Set)$ means that the canonical inclusion functor $\iota:Mod_{\mathcal{T}}\longrightarrow Fun(\mathcal{T},Set)$ admits a reflection, that is a left adjoint functor $r:Fun(\mathcal{T},Set)\longrightarrow Mod_{\mathcal{T}}$.

In the textbook I am reading (Borceux, Handbook of Categorical Algebra, vol. 2, pag. 138) the proof of this fact depends on the so called adjoint functor theorem. I have understood all the conditions of the theorem, but one: the soluton set condition (ssc).

In this case, the ssc says: let $F:\mathcal{T}\longrightarrow Set$ be a functor. The inclusion $\iota:Mod_{\mathcal{T}}\longrightarrow Fun(\mathcal{T},Set)$ satisfies the ssc with respect to $F$ if there exists a set $S_F$ of $\mathcal{T}$-models such that:

for all $G\in Mod_{\mathcal{T}}$ for all natural transformations $\phi:F\Longrightarrow \iota(G)$ there exist $G'\in S_F$ and natural transformations $\alpha:G'\Longrightarrow G, \beta: F\Longrightarrow \iota(G')$ such that $\iota(\alpha)\circ\beta'=\beta$

Here is the proof I found in Borceux, which is incomprehensible to me: consider two functors $F,G:\mathcal{T}\longrightarrow Set$ and a natural transformation $\phi:F\Longrightarrow G$; suppose $G$ preserves finite products. Consider the set $X$ of all elements of $G(T^1)$ of the form $\phi_{T^1}(x)$ for some $x\in F(T^1)$. The cardinality of $X$ is bounded by that of $F(T^1)$. Consider the set $Y$ of all elements of $G(T^1)$ of the form $\alpha(x_1,\ldots,x_n)$, for some integer $n$, $n$-ary operation $\alpha$ and elements $x_i\in X$. The cardinality of $Y$ is bounded by that of $\displaystyle\bigsqcup_{n\in\mathbb{N}}\mathcal{T}(T^n,T^1)\times X^n$, thus by the cardinality of $\displaystyle\bigsqcup_{n\in\mathbb{N}}\mathcal{T}(T^n,T^1)\times F(T^1)^n$. By constuction, $Y$ is stable for al $\mathcal{T}$-operations, thus putting

$$H(T^n)=Y^n\subseteq G(T^1)^n\cong G(T^n)$$ defines a subfunctor $H\subseteq G$ which by construction preserves finite products. Observe that $\phi$ factors through $H$, since given $x\in F(T^n)$, one has $$ G(p_i)(\phi_{T^n}(x))=\phi_{T^1}(F(p_i)(x))\in Y$$ (where $p_i$ is the $i$-th projection) and thus $\phi_{{T^n}}(x)\in Y^n$. Therefore one gets a solution set for $F$ by choosing those finite product preserving functors $H:\mathcal{T}\longrightarrow Set$ such that $H(T^1)$ is a subset of some fixed set $Y$ with cardinality less than the cardinality of $\displaystyle\bigsqcup_{n\in\mathbb{N}}\mathcal{T}(T^n,T^1)\times F(T^1)^n$. Once $H(T^1)$ is fixed, there is indeed (up to a canonical choice of products) just a set of possibilities of constructing a product preserving-functor $H$ since, for every morphism $\beta: T^n\longrightarrow T^m$, there is just a set of possible candidates for $H(\beta)$, i.e. the set of mappings $H(T^1)^n\longrightarrow H(T^1)^m$.

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  • $\begingroup$ Why not try to see what happens in a specific example, say, the theory of groups? $\endgroup$ – Zhen Lin Nov 21 '15 at 11:36

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