Homework 4

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 $P:{\mathbf{\text{Set}}}^J\to \mathbf{\text{Set}}$. To each object $F:J\to \mathbf{\text{Set}}$ (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 ${\mathbf{Set}}^J$, 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 ${\mathbf{Set}}^J$. 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 $\eta :\mathrm{\Delta }P\Rightarrow I_{{\mathbf{\text{Set}}}^J}$; 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.

Note: We could prove this without references to elements once we had an element-free characterization of the annihilator.

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 $p:F(\{\bullet \})\to M$, where $F(\{\bullet \})\simeq R$ is the free $R$-module on a singleton set. The composition ${\pi }_N\circ p:F(\{\bullet \})\to M/N$ 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 $M_s\simeq F(X_s)$. 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 $X={\displaystyle \coprod _{s\in S}{X}_s}$.

An informal proof is simply to note that for every set $Y$ we have ${\displaystyle F(Y)\simeq \underset{y\in Y}{⨁}R}$. In particular,

$F(X)\simeq \underset{x\in X}{⨁}R\simeq \underset{s\in S}{⨁}\underset{x\in X_s}{⨁}R\simeq \underset{s\in S}{⨁}{M}_s.$

The only point in the above chain of isomorphisms that is a bit informal is in the second isomorphism, which is really a consequence of the fact that the construction of direct sum (of $R$-modules) commutes with disjoint union (of sets).

For a more formal proof that doesn't rely on that fact, 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 ${\mathrm{Hom}}_{R\text{-}\mathbf{\text{Mod}}}(M,N)\simeq {\mathrm{Hom}}_{\mathbf{\text{Set}}}(X,U(N))$, 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 ${\displaystyle X=\coprod _{s\in S}{X}_s}$.