Homework 3

1. We have seen how to assign to each group $G$ a one-object category $\mathbf{G}$ whose arrows are labeled by the elements of $G$ (and with arrow composition corresponding to the group operation). By construction, the identity element in $G$ is sent to the identity arrow (on the unique object) in the category $\mathbf{G}$.

   Now suppose $C$ is a category with a unique object and in which every arrow is invertible. Let $G$ be the set of arrows in $C$, and define a binary operation on $G$ using the arrow composition in $C$. (Note that every pair of arrows is composable, since all arrows in $C$ start and end at the unique object.) Since $C$ is a category, its composition law is associative and the identity arrow acts as a two-sided identity. It follows that the operation on $G$ is associative and the element corresponding to the identity arrow acts as two-sided identity. By assumption, every arrow in $C$ is invertible, which by construction of $G$ means that every element in $G$ has an inverse under the so-defined binary operation. Thus $G$ is a group.

   Finally, it is "clear" these two processes are mutually inverse. (Here we should technically make a bijection between isomorphism classes of groups and something analogous on the category side.)

2. Suppose $\varphi :G\to H$ is a group morphism. We wish to define a functor $\mathrm{\Phi }:\mathbf{G}\to \mathbf{H}$. Each of the categories $\mathbf{G}$ and $\mathbf{H}$ has a unique object, so the object function of $\mathrm{\Phi }$ must map the unique object of $\mathbf{G}$ to the unique object of $\mathbf{H}$. The set of arrows in $\mathbf{G}$ is $G$ and the set of arrows in $\mathbf{H}$ is $H$, so on arrows we define $\mathrm{\Phi }$ by $\varphi$. By our construction of the categories $\mathbf{G}$ and $\mathbf{H}$, the map $\mathrm{\Phi }:\mathbf{G}\to \mathbf{H}$ sends the (unique) identity arrow in $\mathbf{G}$ to the (unique) identity arrow in $\mathbf{H}$ (because $\varphi$ is a group morphism and hence $\varphi (e_G)=e_H$), and is compatible with composition (since $\varphi$ is a group morphism and hence $\varphi (g_1{g}_2)=\varphi (g_1)\varphi (g_2)$). Thus, $\mathrm{\Phi }$ does indeed define a functor from $\mathbf{G}$ to $\mathbf{H}$.

   Conversely, suppose $\mathrm{\Phi }:C\to D$ is a functor between one-object categories in which every arrow is invertible. Let $G$ and $H$ be the groups corresponding to $C$ and $D$, respectively. Then the set of elements of $G$ (respectively, $H$) is the set of arrows in $C$ (respectively, $D$), so the arrow function of the functor $\mathrm{\Phi }$ is a map $\varphi :G\to H$. The compatibility of the functor $\mathrm{\Phi }$ with arrow composition (and the fact that it sends the identity arrow to the identity arrow) directly translate to $\varphi$ being a group morphism.

   Finally, it is once again "clear" that these two processes are mutually inverse. (Once again, this is not technically true. We are really sketching an equivalence between two specific categories.)

3. Suppose first that $\varphi :G\to S_X$ is a group morphism. We define a functor $F:\mathbf{G}\to \mathbf{\text{Set}}$ as follows. On objects, $F$ maps the unique object of $\mathbf{G}$ to the set $X$. On arrows, $F$ maps each arrow in $\mathbf{G}$ (corresponding to an element $g\in G$) to the corresponding bijection $\varphi (g):X\to X$. (Recall that the group $S_X$ consists of all set bijections from $X$ to itself.) Note that the identity element $e\in G$ must be sent to the identity permutation $1_X:X\to X$, and that the group morphism property of $\varphi$ ensures the functor $F$ is compatible with composition: $F(g_1{g}_2)=\varphi (g_1{g}_2)=\varphi (g_1)\varphi (g_2)=F(g_1)F(g_2)$. We have therefore indeed defined a functor $F:\mathbf{G}\to \mathbf{\text{Set}}$.

   Conversely, suppose $F:\mathbf{G}\to \mathbf{\text{Set}}$ is a functor. The object function of $F$ assigns to the unique object of $\mathbf{G}$ a set $X$. The arrow function of $F$ assigns to each arrow in $\mathbf{G}$ (corresponding to an element $g\in G$) an arrow $\varphi (g):X\to X$, with identity arrow (corresponding to the identity element $e\in G$) sent to the identity arrow $1_X:X\to X$. The compatibility of the functor with composition implies $\varphi (g_1{g}_2)=F(g_1{g}_2)=F(g_1)F(g_2)=\varphi (g_1)\varphi (g_2)$. This implies both that $\varphi$ is has the group morphism property and also that every map $\varphi (g):X\to X$ is invertible (i.e., bijective), since every arrow in $\mathbf{G}$ is invertible. Thus, $\varphi (g)\in S_X$ and $\varphi :G\to S_X$ is a group morphism.

   As with the previous two parts, these two associations are "clearly" mutually inverse. (We're really asking for trouble at this point.)

4. First suppose there is a natural transformation $\tau :\mathrm{\Phi }\Rightarrow \mathrm{\Psi }$ where $\mathrm{\Phi },\mathrm{\Psi }$ are functors from $\mathbf{G}$ to $\mathbf{H}$. For the sake of clarity, let us label the unique object in $\mathbf{G}$ by $a$ and the unique object in $\mathbf{H}$ by $b$. Note that both $\mathrm{\Phi }$ and $\mathrm{\Psi }$ must map $a$ to $b$. The natural transformation $\tau$ is then entirely defined by a single arrow ${\tau }_a:b\to b$ in $\mathbf{H}$, i.e., by an element $h\in H$. The naturality condition requires that for every arrow $g:a\to a$ in $\mathbf{G}$ (corresponding to an element $g\in G$) we have a commutative diagram

   In other words, we must have $h\varphi (g)=\psi (g)h$, i.e., $\psi (g)=h\varphi (g)h^{-1}$. Since this must hold for all $g\in G$, this exactly says $\varphi$ and $\psi$ are conjugate maps.

   Conversely, suppose $\varphi ,\psi :G\to H$ are conjugate group morphisms, so that there exists some $h\in H$ such that $\psi (g)=h\varphi (g)h^{-1}$ for all $g\in G$. Let $\mathrm{\Phi },\mathrm{\Psi }:\mathbf{G}\to \mathbf{H}$ be the functors corresponding to $\varphi$ and $\psi$, respectively. Define a map $\tau :\mathrm{\Phi }\Rightarrow \mathrm{\Psi }$ by sending the unique object $a$ of $\mathbf{G}$ to the arrow ${\tau }_a$ in $\mathbf{H}$ corresponding to the element $h\in H$. The conjugate property exactly ensures that this defines a natural transformation from $\mathrm{\Phi }$ to $\mathrm{\Psi }$, by the same diagram as above.

1. Let's denote the functor we are creating by ${\bullet }^S:\mathbf{\text{Set}}\to \mathbf{\text{Set}}$. It is already defined on objects, so we need to define it on arrows. To that end, let $f:X\to Y$ be a set map. We can define a set map $f^S:X^S\to Y^S$ by sending each map $h:S\to X$ (which is what the elements of the set $X^S$ are) to the map $f\circ h:S\to Y$ (which is an element of the set $Y^S$). In other words, $f^S$ is the "compose with $f$" map. Observe that if $1_X:X\to X$ is an identity arrow in $\mathbf{\text{Set}}$, then $(1_X{)}^S$ is the "compose with $1_X$" map, which is the identity on $X^S$. (Sanity check: for any $h:S\to X$, the composition $1_X\circ h$ is the same set map $h:X\to S$.)

   It remains to check that this map on arrows is compatible with composition, so suppose $f:X\to Y$ and $g:Y\to Z$ are two set maps. For any $h\in X^S$ (i.e., set map $h:S\to X$), observe that $(g\circ f{)}^S(h)=(g\circ f)\circ h=g\circ (f\circ h)=g^S(f^S(h))$, where the second equality held by the associativity of function composition. Thus, we do indeed have $(g\circ f{)}^S=g^S\circ f^S$ and hence our arrow map is compatible with composition. We therefore have defined a functor ${\bullet }^S:\mathbf{\text{Set}}\to \mathbf{\text{Set}}$, as desired.

2. Define the functor ${\bullet }^S\times S:\mathbf{\text{Set}}\to \mathbf{\text{Set}}$ by sending each set $X$ to the set $X^S\times S$, and each set map $f:X\to Y$ to the set map $f^S\times 1_S:X^S\times S\to Y^S\times S$. (At the level of elements this is simply $(f^S\times 1_S)(h,s)=(f\circ h,s)$.) It is straightforward to show this does indeed define a functor, with the verification being nearly identical to the steps used in part 1.

3. Recall that a natural transformation $\tau :{\bullet }^S\times S\Rightarrow I$ is the data consisting of a set map ${\tau }_X:X^S\times S\to X$ for every set $X$, such that for every set map $f:X\to Y$ the diagram below commutes:

   In the case of our evaluation map $e_X:X^S\times S\to X$ we are being asked to show that the diagram below is commutative:

   Let's chase some elements. Starting from the top-left corner, let $(h,s)$ be any element of the set $X^S\times S$. Going across and down corresponds to $f(e_X(h,s))=f(h(s))$, while going down and then across corresponds to $e_y((f^S\times 1_S)(h,s))=e_Y(f\circ h,s)=(f\circ h)(s)$. By the definition of function composition in $\mathbf{\text{Set}}$, we always have $f(h(s))=(f\circ h)(s)$, and so the diagram does indeed commute. Thus, evaluation does indeed define a natural transformation. (Did you ever really doubt it?)

Recall that for each diagram in $\mathbf{\text{Set}}$ of the form

the pullback consists of the set $X{\times }_{Z}Y=\{(x,y)\mid f(x)=g(y)\}$ together with the set maps ${\pi }_1:X{\times }_{Z}Y\to X$ and ${\pi }_2:X{\times }_{Z}Y\to Y$ defined by $(x,y)\mapsto x$ and $(x,y)\mapsto y$, respectively. The set $X{\times }_{Z}Y$ (together with the maps ${\pi }_1$ and ${\pi }_2$) are characterized by a certain universal property (which we'll return to later).

Encoding the diagrams as objects in a functor category

To see how the pullback construction is an adjoint functor, we first encode such diagrams as elements of a functor category. Let $J$ be the category with three objects and two non-identity arrows, visualized as $a\stackrel{\varphi }{\to }c\stackrel{\psi }{\leftarrow }b$. Functors $J\to \mathbf{\text{Set}}$ then correspond to diagrams in $\mathbf{\text{Set}}$ of shape $J$; i.e., diagrams of the form $X\stackrel{f}{\to }Z\stackrel{g}{\leftarrow }Y$ in $\mathbf{\text{Set}}$, exactly as shown above. (Here we are writing $X$, $Y$, and $Z$ for $F(a)$, $F(b)$, and $F(c)$, respectively; we are also writing $f$ and $g$ for $F(\varphi )$ and $F(\psi )$. This is mainly to cut down on the notation.)

Let ${\mathbf{\text{Set}}}^J$ denote the category of functors $J\to \mathbf{Set}$. Note that the arrows in ${\mathbf{\text{Set}}}^J$ are natural transformations $\tau :F\Rightarrow G$, which correspond to to triples of set maps ${\tau }_a:F(a)\to G(a)$, ${\tau }_b:F(b)\to G(b)$, ${\tau }_c:F(c)\to G(c)$ such that the diagram below commutes. (Once again, we are writing $X$ for $F(a)$, $X^{\prime }$ for $G(a)$, etc.)

Defining the pullback functor

We now define a functor $P:{\mathbf{\text{Set}}}^{\mathcal{J}}\to \mathbf{\text{Set}}$. To each object $F:\mathcal{J}\to \mathbf{\text{Set}}$ (which corresponds to a diagram $X\stackrel{f}{\to }Z\stackrel{g}{\leftarrow }Y$) we assign $P(F)=X{\times }_{Z}Y$. To each arrow $\tau :F\Rightarrow G$ we define the set map $P(\tau ):X{\times }_{Z}Y\to X^{\prime }{\times }_{Z^{\prime }}{Y}^{\prime }$ by $(x,y)\mapsto ({\tau }_a(x),{\tau }_c(y))$. Note that the commutativity of the above diagrams ensures that this map is well defined. Each pair $(x,y)\in X{\times }_{Z}Y$ satisfies $f(x)=g(y)$ and hence ${\tau }_c(f(x))={\tau }_c(g(y))$. Commutativity of the above diagram then gives

$f^{\prime }({\tau }_a(x))={\tau }_c(f(x))={\tau }_c(g(y))=g^{\prime }({\tau }_b(y))$

and hence the pair $({\tau }_a(x),{\tau }_b(y))$ is indeed an element of $X^{\prime }{\times }_{Z^{\prime }}{Y}^{\prime }$.

(Optional) Verifying $P$ is a functor

We should verify that we have actually defined a functor $P$. First note that if $1_F:F\Rightarrow F$ is an identity arrow in ${\mathbf{Set}}^{\mathcal{J}}$, then ${\tau }_a=1_X$ and ${\tau }_b=1_Y$ are identity set maps, hence $P(1_F)$ is the identity set map on $X{\times }_{Z}Y$. Next suppose $\tau :F\Rightarrow G$ and $\eta :G\Rightarrow H$ are a pair of composable arrows in ${\mathbf{Set}}^{\mathcal{J}}$. This corresponds to a commutative diagram

Then $P(\eta \circ \tau ):X{\times }_{Z}Y\to X^{″}{\times }_{Z^{″}}{Y}^{″}$ is the set map given by $(x,y)\mapsto ({\eta }_a{\tau }_a(x),{\eta }_b{\tau }_b(y))$, and this is easily seen to be the same as the set map $P(\eta )\circ P(\tau )$.

Showing the pullback functor is right adjoint to the diagonal functor

Defining the set maps

So we do indeed have a functor $P:{\mathbf{Set}}^J\to \mathbf{Set}$ that sends each diagram $X\stackrel{f}{\to }Z\stackrel{g}{\leftarrow }Y$ in $\mathbf{Set}$ to the pullback set $X{\times }_{Z}Y$ of that diagram. We will show this functor is right adjoint to the diagonal functor $\mathrm{\Delta }:\mathbf{Set}\to {\mathbf{Set}}^J$, which sends each set $C$ to the "constant diagram" $C\stackrel{1_C}{\to }C\stackrel{1_C}{\leftarrow }C$. To that end, we must establish a natural bijection

${\varphi }_{C,F}:{\mathrm{Hom}}_{{\mathbf{Set}}^J}(\mathrm{\Delta }C,F)\stackrel{\sim }{\to }{\mathrm{Hom}}_{\mathbf{Set}}(C,P(F)).$

For each fixed set $C$ and diagram $F$ (corresponding to $X\stackrel{f}{\to }Z\stackrel{g}{\leftarrow }Y$), natural transformations $\tau :\mathrm{\Delta }C\to F$ correspond to triples of set maps ${\tau }_a:C\to X$, ${\tau }_b:C\to Y$, ${\tau }_c:C\to Z$ giving commutative diagrams

We can then define a set map ${\varphi }_{C,F}(\tau ):C\to X{\times }_{Z}Y$ by $c\mapsto ({\tau }_a(c),{\tau }_b(c))$; the commutativity of the above diagram ensures the pair $({\tau }_a(c),{\tau }_b(c))$ really is in the set $X{\times }_{Z}Y$.

(Optional) Verifying the set maps are bijections

We next verify that ${\varphi }_{C,F}$ is a set bijection by defining the inverse set map ${\psi }_{C,F}$. Take any set map $h:C\to P(F)$, i.e., set map $h:C\to X{\times }_{Z}Y$. We then have a commutative diagram

Let ${\tau }_a={\pi }_1\circ h$, ${\tau }_b={\pi }_2\circ g$, and ${\tau }_c=f\circ {\pi }_1\circ h=g\circ {\pi }_2\circ h$. We then have a commutative diagram (with ${\tau }_c$ not explicitly drawn, to keep the diagram as clean as possible):

This shows we have defined a natural transformation $\tau :\mathrm{\Delta }C\Rightarrow F$. Let ${\psi }_{C,F}(h)=\tau$.

We claim ${\varphi }_{C,F}$ and ${\psi }_{C,F}$ are mutual inverses. First suppose $\tau :\mathrm{\Delta }C\to F$ is a natural transformation, corresponding to the commutative diagram

Then ${\varphi }_{C,F}(\tau )$ is the map $h:C\to X{\times }_{Z}Y$ defined by $c\mapsto ({\tau }_a(c),{\tau }_b(c))$. We then have a commutative diagram

Although not drawn, note that we have ${\tau }_c=f\circ {\pi }_1\circ h=g\circ {\pi }_2\circ h$. It follows that the associated natural transformation ${\psi }_{C,F}(h):\mathrm{\Delta }C\Rightarrow F$ is exactly $\tau$. In other words, ${\psi }_{C,F}({\varphi }_{C,F}(\tau ))=\tau$. This proves ${\psi }_{C,F}\circ {\varphi }_{C,F}$ is the identity on the set of natural transformations $\mathrm{\Delta }C\Rightarrow F$.

Conversely, suppose $h:C\to X{\times }_{Z}Y$ is a set map. Then the associated natural transformation $\tau ={\psi }_{C,F}(h)$ is defined so that we have a commutative diagram:

For reference, ${\tau }_a={\pi }_1\circ h$, ${\tau }_b={\pi }_2\circ g$, and ${\tau }_c=f\circ {\pi }_1\circ h=g\circ {\pi }_2\circ h$. In any case, note that by construction we have $h(c)=({\tau }_a(c),{\tau }_b(c))$, which is precisely the definition of ${\varphi }_{C,F}(\tau )$. Thus, we do indeed have ${\varphi }_{C,F}({\psi }_{C,F}(h))=h$ and hence ${\varphi }_{C,F}\circ {\psi }_{C,F}$ is the identity on the set of maps $C\to X{\times }_{Z}Y$.

(Optional) Verifying the sets maps are natural

It remains to verify ${\varphi }_{C,F}$ is a natural bijection. We'll first show it's natural in the set $C$. Suppose $p:C\to C^{\prime }$ is a set map. We need to verify that the diagram below is commutative:

Take an element ${\tau }^{\prime }:\mathrm{\Delta }{C}^{\prime }\Rightarrow F$ from the set on the bottom left. That corresponds to a commutative diagram

The left vertical map sends the natural transformation ${\tau }^{\prime }$ to the natural transformation $\tau :\mathrm{\Delta }C\to F$ given by the commutative diagram below:

In other words, ${\tau }_a={\tau }_a^{\prime }\circ p$, ${\tau }_b={\tau }_b^{\prime }\circ p$, and ${\tau }_c={\tau }_c^{\prime }\circ p$. The map ${\varphi }_{C,F}$ then sends $\tau$ to the set map $h:C\to X{\times }_{Z}Y$ defined by $c\mapsto ({\tau }_a(c),{\tau }_b(c))=({\tau }_a^{\prime }(p(c)),{\tau }_b^{\prime }(p(c)))$.

On the other hand, the bottom horizontal map ${\varphi }_{C^{\prime },F}$ sends the natural transformation ${\tau }^{\prime }$ to the map $h^{\prime }:C^{\prime }\to X{\times }_{Z}Y$ defined by $c^{\prime }\mapsto ({\tau }_a^{\prime }(c^{\prime }),{\tau }_b^{\prime }(c^{\prime }))$. The right vertical map then sends this to the set map $h^{\prime }\circ p:C\to X{\times }_{Z}Y$ defined by $c\mapsto ({\tau }_a^{\prime }(p(c)),{\tau }_b^{\prime }(p(c)))$. This is exactly the same as the map we found above (by going up and across). Thus, we do indeed have a commutative diagram.

Finally, we will verify ${\varphi }_{C,F}$ is natural in $F$. So suppose $\alpha :F\Rightarrow F^{\prime }$ is a natural transformation. We need to show the diagram below is commutative:

Take an element $\tau :\mathrm{\Delta }C\Rightarrow F$ in the top-left set. This corresponds to a commutative diagram

The left vertical map sends this to the natural transformation ${\tau }^{\prime }=\alpha \circ \tau$ as shown in the commutative diagram below:

In other words, ${\tau }_a^{\prime }={\alpha }_a\circ {\tau }_a$, ${\tau }_b^{\prime }={\alpha }_b\circ {\tau }_b$, and ${\tau }_c^{\prime }={\alpha }_c\circ {\tau }_c$. The map ${\varphi }_{C,F^{\prime }}$ then sends ${\tau }^{\prime }$ to the set map $h^{\prime }:C\to X^{\prime }{\times }_{Z^{\prime }}{Y}^{\prime }$ defined by $c\mapsto ({\tau }_a^{\prime }(c),{\tau }_b^{\prime }(c))=({\alpha }_a({\tau }_a(c)),{\alpha }_b({\tau }_b(c)))$.

On the other hand, the top horizontal map ${\varphi }_{C,F}$ sends $\tau$ to the set map $h:C\to X{\times }_{Z}Y$ defined by $c\mapsto ({\tau }_a(c),{\tau }_b(c))$. The right vertical map then sends $h$ to the set map $P(\alpha )h:C\to X^{\prime }{\times }_{Z^{\prime }}{Y}^{\prime }$ defined by $c\mapsto ({\alpha }_a({\tau }_a(c)),{\alpha }_b({\tau }_b(c)))$. This is exactly the same as the set map $h^{\prime }$ we found above. Thus, we do indeed have another commutative diagram.

The unit and counit of the adjunction

The unit of our adjunction is a natural transformation $\tau :I_{\mathbf{\text{Set}}}\Rightarrow P\mathrm{\Delta }$; i.e., a family of set maps ${\tau }_C:C\to C{\times }_{C}C$ that is natural in $C$. Based on all of our work above, these maps are given by $c\mapsto (c,c)$.

The counit of our adjunction is a natural transformation $\eta :\mathrm{\Delta }P\Rightarrow I_{{\mathbf{\text{Set}}}^{\mathcal{J}}}$; i.e., a family of natural transformations ${\eta }_F:\mathrm{\Delta }(P(F))\Rightarrow F$ that is natural in diagrams $F$. Based on our work above, these maps are given by the commutative diagrams below, where as usual the diagram $F$ is labeled in the "standard" way and where ${\pi }_3=f\circ {\pi }_1=g\circ {\pi }_2$:

Suppose $m_1,\dots ,m_k$ generate $M$ as an $R$-module. Since $M$ is torsion, for each $m_i$ there is some nonzero $r_i\in R$ such that $r_i{m}_i=0_M$. Let $r=r_1\cdots r_k$. Note that $r$ is nonzero, since all $r_i$ are nonzero and $R$ is an integral domain. Also note that $r{m}_i=0_M$ for every $i$, since $R$ is commutative.

Now take any $m\in M$ and write $m=s_1{m}_1+\cdots +s_k{m}_k$ for some $s_i\in R$. Then observe that for every $i$ we have $r(s_i{m}_i)=s_i(r{m}_i)=s_i(0_M)=0_M$, once again using the commutativity of $R$. It follows that $rm=0_M$. Since $m\in M$ was arbitrary, this proves that $r$ annihilates $M$. As $r$ is nonzero, this proves the annihilator of $M$ in $R$ is nontrivial.

Suppose $M$ is a cyclic $R$-module and $N\subseteq M$ is a submodule. Let ${\pi }_N:M\to M/N$ be the canonical (surjective) morphism, given by $m\mapsto m+N$. Since $M$ is cyclic, there exists a surjective $R$-module morphism $p:R\to M$. The composition ${\pi }_N\circ p:R\to M/N$ is then also a surjective $R$-module morphism. Thus, $M/N$ is cyclic.

Suppose $m_1+N,\dots ,m_k+N$ generate $M/N$ and $n_1,\dots ,n_l$ generate $N$. We claim that $\{m_1,\dots ,m_k,n_1,\dots ,n_l\}$ is a set that generates $M$. To prove this, take any $m\in M$ and first consider the element $m+N$ in $M/N$. We can write $m+N=r_1(m_1+N)+\cdots +r_k(m_k+N)$ for some $r_i\in R$. This means we have $m+N=(r_1{m}_1+\cdots +r_k{m}_k)+N$, and hence $m-(r_1{m}_1+\cdots +r_k{m}_k)\in N$. We can now write this difference using the generators for $N$, so that

${\displaystyle m-(r_1{m}_1+\cdots +r_k{m}_k)=s_1{n}_1+\cdots +s_l{n}_l}$

for some $s_j\in R$. But we now have

$m=r_1{m}_1+\cdots +r_k{m}_k+s_1{n}_1+\cdots s_l{n}_l$

As $m\in M$ was arbitrary, this proves the set $\{m_1,\dots ,m_k,n_1,\dots ,n_l\}$ generates $M$.

Suppose $\{M_s\mid s\in S\}$ is a family of free $R$-modules. Then for each $s\in S$ there is some set $X_s$ such that $M\simeq F(X_s)$. Let $M={\displaystyle \underset{s\in S}{⨁}{M}_s}$, with inclusion morphisms $j_s:M_s\to M$. We claim $M\simeq F(X)$, where $X={\displaystyle \coprod _{s\in S}{X}_s}$.

To prove this, let $f:M\to N$ be any $R$-module morphism. For each $s\in S$ the composition $f\circ j_s:M_s\to N$ is then an $R$-module morphism, and since $M_s\simeq F(X_s)$ this corresponds (by the universal property of $F(X_s)$) to a set map $g_s:X_s\to U(N)$ (where $U$ is the usual forgetful functor from $R$-modules to sets). By the universal property of the coproduct (which is just the disjoint union in $\mathbf{\text{Set}}$) we then have a set map $g:{\displaystyle \coprod _{s\in S}{X}_s\to U(N)}$.

Conversely, suppose $g:{\displaystyle \coprod _{s\in S}{X}_s\to U(N)}$ is any set map. By the universal property of the direct product we then have set maps $g_s:X_s\to U(N)$ for every $s\in S$, which in turn correspond uniquely to $R$-module morphisms $f_s:F(X_s)\to N$. By the universal property of direct sum this family of morphisms corresponds uniquely to an $R$-module morphism $f:{\displaystyle \underset{s\in S}{⨁}F(X_s)\to N}$. Pre-composing with the isomorphism $M\simeq {\displaystyle \underset{s\in S}{⨁}F(X_s)}$ yields an $R$-module morphism $\tilde{f}:M\to N$.

We've thus established a (natural, although we didn't actually prove naturality) bijection ${\mathrm{Hom}}_R(M,N)\simeq {\mathrm{Hom}}_{\mathbf{\text{Set}}}(X,U(N))$, and thus $M\simeq F(X)$, as desired.

In much less fancy language, if in each $M_s$ every element can be expressed uniquely as an $R$-linear sum of elements in $X_s$, then in $M$ each element can be expressed uniquely as an $R$-linear sum of elements in ${\displaystyle X=\coprod _{s\in S}{X}_s}$.