Proof

Let $V_1, V_2 \colon \mathcal{D} \to Set$ be two product-preserving covariant functors. From the definition of a multi-sorted infinitary category we see that we may choose a set of sorts, say $S$. Let $D_s \in \mathcal{D}$ be the image of $s \in S$.

As $S$ is a set, the product

is a set. We define a function $[V_1,V_2] \to \prod_{s \in S} Set(V_1(D_s), V_2(D_s))$ by sending a natural transformation $\alpha \colon V_1 \to V_2$ to the product of its values at each $D_s$. That is,

Let $D \in \mathcal{D}$ be an arbitrary object. From the definition of a multi-sorted infinitary Lawvere theory, there is an isomorphism

for some sets $X_s$. For each $s \in S$ and $x \in X_s$, there is a projection $p_{s,x} \colon \prod_{s \in S} D_s^{X_s} \to D_s$. As $V_1$ and $V_2$ are product-preserving, we have isomorphisms

commuting with the projections. Thus for each $s \in S$ and $x \in X_s$, we have the following commutative diagram.

Since this holds for all $s \in S$ and $x \in X_s$, by the basic properties of products, there is a unique morphism $\prod_{s \in S} V_1(D_s)^{X_s} \to \prod_{s \in S} V_2(D_s)^{X_s}$ making the diagram commute. This morphism is normally written $\prod_{s \in S} \alpha_{D_s}^{X_s}$. Thus under the isomorphism $d$ above, $\alpha_D$ is taken to $\prod_{s \in S} \alpha_{D_s}^{X_s}$. As this holds for all $D \in \mathcal{D}$, $\alpha$ is completely determined by the $\alpha_{D_s}$, whence the map in (1) is injective.

Hence $[V_1,V_2]$ is a subset of a set and thus a set.

Proof

To ease the notation, let us write $U$ for the forgetful (“underlying set(s)”) functor and $F$ for the free functor.

To show the adjunction, we define the unit and counit natural transformations:

Let us consider $\eta$. To define this, we evaluate it at an $S$-graded set, $X$, to get a morphism of graded sets, $\eta_X$. Such a morphism is itself a natural transformation so we evaluate again at $s_0 \in S$. At this point, we have a function (in $Set$)

This is simple to describe: it is assignment to $x \in X_{s_0}$ of the projection onto the $x$th $D_{s_0}$-factor:

The counit, $\epsilon$, is a little more complicated to describe. Let $V \colon \mathcal{D} \to Set$ be a covariant product-preserving functor. Evaluated at $V$, $\epsilon_V$ is a natural transformation

By the Yoneda lemma, such natural transformations are in bijective correspondence with elements of $V\left(\prod_{s \in S} D_s^{V(D_s)}\right)$. As $V$ is product-preserving, there is a natural isomorphism:

Now for any set $Z$, there is an obvious element in $Z^Z$: the one that has a $z$ in the $z$th place (which corresponds to the identity function $Z \to Z$). Taking the product of these gives an element in $\prod_{s \in S} V(D_s)^{V(D_s)}$ which we use to define $\epsilon_V$.

Let us now prove that these provide the desired adjunction. For one half, we need to consider the composition:

Let us apply this to a graded set $X = (s \mapsto X_s)$. Then we are looking at a natural transformation from

to itself. By the standard Yoneda argument, it is sufficient to look at the induced function on

and in particular at the image of the identity therein.

The first part comes from $F\eta$ at $F(X)$. For each $s_0 \in S$, we have a function $\eta_{X,s_0} \colon X_{s_0} \to \mathcal{D}\left(\prod_{s \in S} D_s^{X_s}, D_{s_0}\right)$ and thus a morphism

What this does is the following: it sends component corresponding to the $(s,x)$th projection to the $x$th component and all other components are forgotten. Thus we have a morphism:

Under this, the identity morphism goes to the projection morphism described just above.

Now we apply $\epsilon F$ to get a morphism:

To find this element, we consider the diagram:

In this diagram, we have left off the subscript on $\epsilon$ for conciseness. The vertical morphism is that induced by the projection from (2). Since we want to have this projection itself in the lower-left, an obvious place to start is with the identity in the top-left. The right-hand square commutes since the vertical maps are projections. Starting with the identity in the top-left, we get the “Yoneda element” corresponding to $\epsilon$ in the top-right. That element can be written $(f)_f$. The vertical map selects the $p_{s,x}$th element of the list and puts it in the $(s,x)$th slot. As this is the projection $p_{s,x}$, when moving back to the lower-middle, we obtain the identity morphism as required.

Now let us turn to the other half. We need to consider the composition:

So we need to start with a covariant product-preserving functor $V \colon \mathcal{D} \to Set$ and apply $U$. Then we are looking at a morphism of graded sets from

to itself. So we pick $s_0 \in S$ and look at the function

We start with the function defined by the unit $\eta$:

which sends $v \in V(D_{s_0})$ to the projection $\prod_{s \in S} D_s^{V(D_s)} \to D_{s_0}$ which corresponds to the $v$th factor of the $s_0$th factor.

Now we apply $\epsilon$. This gives us a natural transformation

which we evaluate at $D_{s_0}$ (technically, which we evaluate at $(s \mapsto D_s)$ which we then evaluate at $s_0$). This natural transformation is determined by an element of $V\left(\prod_{s \in S} D_s^{V(D_s)}\right)$ and the element that we want is the image of that element under the function induced by the projection:

As $V$ is product-preserving, we can rearrange this to:

The projection selects the $(s_0,v)$th term of the element in question, and this element is, by definition, $v$. Hence the induced function on $V(D_{s_0})$ is the identity and the adjunction is shown.