Midterm Exam Solutions

We have two options:

A very short proof

The shortest solution is to recall that if $X=\{x_1,\dots ,x_n\}$ is a set with $n$ elements, then we have $R$-module isomorphisms

$$F(X)\simeq \underset{n\text{ times}}{\underbrace{R\oplus \cdots \oplus R}}=\underset{i=1}{\overset{n}{⨁}}R,$$

and so (using another problem from the study guide with $M=R$) we have

$${\mathrm{Hom}}_R(F(X),R)\simeq {\mathrm{Hom}}_R(\underset{i=1}{\overset{n}{⨁}}R,R)\simeq \underset{i=1}{\overset{n}{⨁}}{\mathrm{Hom}}_R(R,R)\simeq \underset{i=1}{\overset{n}{⨁}}R\simeq F(X).$$

If we use this solution to part (1), then for part (2) we need to chain together the actual maps behind the above isomorphisms. When we do so, we'll arrive at the same map described below.

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A direct proof

We can also give a direct proof. Define a map $\varphi :{\mathrm{Hom}}_R(F(X),R)\to F(X)$ by sending each $R$-module morphism $f:F(X)\to R$ to the element $\varphi (f)={\displaystyle \sum _{i=1}^{n}f(x_i)\cdot x_i}$. (Here we are identifying $x_i$ with its image in $F(X)$. Under the isomorphism $F(X)\simeq \underset{i}{⨁}R$ the corresponding element is what we have previously called $e_i$, the formal sum in which all elements are $0_R$ except the $i^{\text{th}}$ element, which is $1_R$.) This map is additive:

$$\begin{array}{rl}\varphi (f_1+f_2)& =\sum _{i=1}^n(f_1+f_2)(x_i)\cdot x_i\\ & =\sum _{i=1}^n(f_1(x_i)+f_2(x_i))\cdot x_i\\ & =\sum _{i=1}^n{f}_1(x_i)\cdot x_i+\sum _{i=1}^n{f}_2(x_i)\cdot x_i\\ & =\varphi (f_1)+\varphi (f_2).\end{array}$$

It is also compatible with the left $R$-action:

$$\varphi (r\cdot f)=\sum _{i=1}^n(r\cdot f)(x_i)\cdot x_i=\sum _i^{n}rf(x_i)\cdot x_i=r\cdot \sum _{i=1}^{n}f(x_i)\cdot x_i=r\cdot \varphi (f).$$

It is therefore an $R$-module morphism.

As for the inverse map, define $\psi :F(X)\to {\mathrm{Hom}}_R(F(X),R)$ by sending each formal sum $s={\displaystyle \sum _{i=1}^n{r}_i{x}_i}$ to the map ${\psi }_s:F(X)\to R$ defined by ${\psi }_s(x_i)=r_i$. We automatically have that ${\psi }_s$ is an $R$-module morphism and hence an element of ${\mathrm{Hom}}_R(F(X),R)$. Moreover, the map $\psi$ we have defined is indeed additive. If $s={\displaystyle \sum _{i=1}^n{r}_i{x}_i}$ and $s^{\prime }={\displaystyle \sum _{i=1}^n{r}_i^{\prime }{x}_i}$, then $s+t={\displaystyle \sum _{i=1}^n(r_i+r_i^{\prime })x_i}$ and for all $i$ we have

$${\psi }_{s+s^{\prime }}(x_i)=r_i+r_i^{\prime }={\varphi }_s(x_i)+{\varphi }_{s^{\prime }}(x_i),$$

hence ${\psi }_{s+s^{\prime }}={\psi }_s+{\psi }_{s^{\prime }}$. It is also an $R$-module morphism, since if $s={\displaystyle \sum _{i=1}^n{r}_i{x}_i}$ then $r\cdot s=r\cdot {\displaystyle \sum _{i=1}^n{r}_i{x}_i=\sum _{i=1}^n(r{r}_i)x_i}$ and so for all $i$ we have

$${\psi }_{r\cdot s}(x_i)=r{r}_i=r\cdot {\psi }_s(x_i),$$

hence ${\psi }_{r\cdot s}=r\cdot {\psi }_s$. Thus, $\psi$ is indeed an $R$-module morphism.

Finally, observe that $\varphi$ and $\psi$ are mutual inverses. For each $f\in {\mathrm{Hom}}_R(F(X),R),$ let $s={\displaystyle \sum _{i=1}^{n}f(x_i)x_i}$. Then

$$\psi (\varphi (f))={\displaystyle \psi (\sum _{i=1}^{n}f(x_i)x_i)={\psi }_s,}$$

and for every $i$ we have ${\psi }_s(x_i)=f(x_i)$; i.e., ${\psi }_s=f$. In other words, for every $f$ we have $\psi (\varphi (f))=f$.

Conversely, suppose $s\in F(X)$, say $s={\displaystyle \sum _{i=1}^n{r}_i{x}_i}$. Then ${\psi }_s$ is the map defined by ${\psi }_s(x_i)=r_i$, and so $\varphi ({\psi }_s)={\displaystyle \sum _{i=1}^n{r}_i{x}_i=s}$; i.e., $\varphi ({\psi }_s)=s$. Thus $\varphi$ and $\psi$ are mutual inverses and hence isomorphisms.

1. Define a set map $g:M\times N\to M{\otimes }_{S}P$ by $(m,n)\mapsto m\otimes f(n)$. This map is linear in $M$:

   $$\begin{array}{rl}g(m_1+m_2,n)& =(m_1+m_2)\otimes f(n)\\ & =m_1\otimes f(n)+m_2\otimes f(n)\\ & g(m_1,n)+g(m_2,n).\end{array}$$

   It is linear in $N$, since $f$ is additive:

   $$\begin{array}{rl}g(m,n_1+n_2)& =m\otimes f(n_1+n_2)\\ & =m\otimes (f(n_1)+f(n_2))\\ & =m\otimes f(n_1)+m\otimes f(n_2)\\ & =g(m,n_1)+g(m,n_2).\end{array}$$

   It is $S$-balanced, since $f$ is compatible with the left $S$-actions:

   $$\begin{array}{rl}g(ms,n)& =ms\otimes f(n)\\ & =m\otimes f(sn)\\ & =g(m,sn).\end{array}$$

   It is compatible with the left $R$-actions:

   $$\begin{array}{rl}g(rm,n)& =rm\otimes f(n)\\ & =r\cdot (m\otimes f(n))\\ & =r\cdot g(m,n).\end{array}$$

   It is compatible with the right $T$-actions, since $f$ is compatible with the right $T$-actions:

   $$\begin{array}{rl}g(m,nt)& =m\otimes f(nt)\\ & =m\otimes f(n)t\\ & =(m\otimes f(n))\cdot t\\ & =g(m,n)\cdot t.\end{array}$$

   So, by the universal property of the tensor product the corresponding map $m\otimes n\mapsto m\otimes f(n)$ is a well-defined $(R,T)$-bimodule morphism.

2. Since $f$ is surjective, every element in $P$ is of the form $f(n)$ for some $n\in N$. Since the simple tensors of the form $m\otimes p$ generate $M{\otimes }_{S}P$ as an $(R,T)$-bimodule, it follows that the simple tensors of the form $m\otimes f(n)$ also generate $M{\otimes }_{S}P$ as an $(R,T)$-bimodule. But this implies that the elements $(1_M\otimes f)(m\otimes n)$ generate $M{\otimes }_{S}P$ as an $(R,T)$-bimodule. Since those same elements generate the image of $1_M\otimes f$, this proves the image of $1_M\otimes f$ is all of $M{\otimes }_{S}P$; i.e., the morphism $1_M\otimes f$ is surjective.

1. First consider simple tensors $v\otimes w$. We can write $w={\displaystyle \sum _{i=1}^n{d}_i{e}_i}$ and then use the bilinearity properties of the tensor product to write

   $$v\otimes w=v\otimes \sum _{i=1}^n{d}_i{e}_i=\sum _{i=1}^n(d_{i}v)\otimes e_i.$$

   This shows it is possible to express every simple tensor $v\otimes w$ in the desired form. For a general tensor ${\displaystyle \sum _{j=1}^m{v}_j\otimes w_j}$ we can use the above process to write each summand $v_j\otimes w_j$ in the desired form, and then properties of the tensor product allow us to write the entire tensor in the desired form:

   $$\sum _{j=1}^m{v}_j\otimes w_j=\sum _{j=1}^m\sum _{i=1}^n{v}_{ji}\otimes e_i=\sum _{i=1}^n\left(\sum _{j=1}^m{v}_{ji}\right)\otimes e_i.$$

   This proves that every tensor can be written in the desired form.

   It remains to prove uniqueness of each such expression. To that end, suppose $v\otimes w$ can be written both as ${\displaystyle \sum _{i=1}^n{v}_i\otimes e_i}$ and as ${\displaystyle \sum _{i=1}^n{v}_i^{\prime }\otimes e_i}$ for some vectors $v_i,v_i^{\prime }\in V$. We then have

   $$\sum _{i=1}^n(v_i-v_i^{\prime })\otimes e_i=0.$$

   It now suffices to prove that ${\displaystyle \sum _{i=1}^n{u}_i\otimes e_i=0}$ exactly when $u_i=0$ for all $i$. For this, write each $u_i$ in terms of the basis as $u_i={\displaystyle \sum _{j=1}^n{c}_{ij}{e}_j}$. We then have

   $$0=\sum _{i=1}^n{u}_i\otimes e_i=\sum _{i=1}^n\left(\sum _{j=1}^n{c}_{ij}{e}_j\right)\otimes e_i=\sum _{1\le i,j\le n}{c}_{ij}(e_j\otimes e_i).$$

   But the set $\{e_j\otimes e_i\mid 1\le i,j\le n\}$ is a basis for $V{\otimes }_{F}V$, hence linearly independent, so the above equality implies all $c_{ij}=0$ and hence all $u_i=0$.

2. First note that if $v,w\in V$ are such that $w=av$ for some $a\in F$ then we immediately have $v\otimes w=v\otimes (av)=(av)\otimes v=w\otimes v$.

   Conversely, suppose $v\otimes w=w\otimes v$. Write $v=c_1{e}_1+\cdots +c_n{e}_n$ and $w=d_1{e}_1+\cdots +d_n{e}_n$. Then observe that

   $$\begin{array}{rl}v\otimes w& =v\otimes \left(\sum _{i=1}^n{d}_i{e}_i\right)\\ & =\sum _{i=1}^n(d_{i}v)\otimes e_i\end{array}$$

   but also

   $$\begin{array}{rl}v\otimes w& =w\otimes v\\ & =w\otimes \left(\sum _{i=1}^n{c}_i{e}_i\right)\\ & =\sum _{i=1}^n(c_{i}w)\otimes e_i\end{array}$$

   By uniqueness of such expressions, we can thus conclude that we must have $d_{i}v=c_{i}w$ for every index $i$. Since $w$ is a nonzero vector we must have $d_i\ne 0$ for at least one index $i$, and for any such index $i$ we must have $v=(d_i^{-1}{c}_i)w$,.

First suppose $\tau$ is a natural isomorphism and let $\eta :G\Rightarrow F$ be the inverse natural transformation. Then for every object $c\in C$ the set maps ${\tau }_c:F(c)\to G(c)$ and ${\eta }_c:G(c)\to F(c)$ satisfy $(\eta \circ \tau {)}_c=1_{F(c)}$ and $(\tau \circ \eta {)}_c=1_{G(c)}$. By the definition of composition for natural transformations, these exactly say that ${\eta }_c\circ {\tau }_c=1_{F(c)}$ and ${\tau }_c\circ {\eta }_c=1_{G(c)}$. In other words, ${\tau }_c$ and ${\eta }_c$ are mutually inverse set maps between the sets $F(c)$ and $G(c)$, and hence both are bijections. In particular, $\tau$ is a natural bijection.

Conversely, suppose $\tau$ is a natural bijection. For each object $c\in C$ define ${\eta }_c:G(c)\to F(c)$ to be the inverse set map to ${\tau }_c$. In other words, for every $x\in F(c)$ and $y\in G(c)$ we have ${\tau }_c(x)=y$ if and only if $x={\eta }_c(y)$. Note that by construction we have ${\eta }_c\circ {\tau }_c=1_{F(c)}$ and ${\tau }_c\circ {\eta }_c=1_{G(c)}$, so assuming $\eta$ is actually a natural transformation it will be inverse to $\tau$ and hence both will be natural isomorphisms. To verify the naturality of $\eta$, suppose $f:c\to c^{\prime }$ is an arrow in $C$. Since $\tau$ is natural, we have a commutative diagram

We claim the diagram below is also commutative

Take an element $y$ in the top-right corner; i.e., an element $y\in G(c)$. Let $x={\eta }_c(y)$. Then mapping across the top horizontal arrow sends $y$ to $x$, and mapping then down the left vertical arrow sends $x$ to $F(f)(x)$. On the other hand, sending $y$ down the right vertical arrow maps $y$ to $G(f)(y)$, and then sending that along the bottom horizontal arrows sends $G(f)(y)$ to ${\eta }_{c^{\prime }}(G(f)(y)).$ We are claiming that

$${\eta }_{c^{\prime }}(G(f)(y))=F(f)(x).$$

Since ${\eta }_{c^{\prime }}$ is the inverse set map to ${\tau }_{c^{\prime }}$, this is equivalent to the condition that

$$G(f)(y)={\tau }_{c^{\prime }}(F(f)(x)).$$

But $y={\tau }_c(x)$, and so the above equality is equivalent to the condition

$$G(f)({\tau }_c(x))={\tau }_{c^{\prime }}(F(f)(x)),$$

and this is true by the naturality of $\tau$. Thus, our maps ${\eta }_c:G(c)\to F(c)$ do indeed define a natural transformation $\eta :G\Rightarrow F$. By construction, $\tau$ and $\eta$ are inverse, hence both are natural isomorphisms.

1. First consider the natural isomorphism ${\alpha }^{-1}\circ \beta :H_d\stackrel{\sim }{\to }{H}_c.$ By Yoneda's Lemma, this corresponds to a unique element of the set $H_d(c)={\mathrm{Hom}}_C(c,d)$; i.e., a unique arrow $f:c\to d$. Under this correspondence, the natural transformation ${\alpha }^{-1}\circ \beta$ is $H(f)$, the "precompose with f" natural transformation. In other words, ${\alpha }^{-1}\circ \beta =H(f)$ and hence $\beta =\alpha \circ H(f)$.

   To prove the uniqueness, suppose $f^{\prime }:c\to d$ is another arrow such that $\beta =\alpha \circ H(f^{\prime })$. Since $\alpha$ is a natural isomorphism, we must then have $H(f^{\prime })={\alpha }^{-1}\circ \beta =H(f).$ Yoneda's Lemma then implies $f^{\prime }=f$.

2. Switching the roles of $c$ and $d$ in the previous part, we see that there is also a unique arrow $g:d\to c$ such that $\alpha =\beta \circ H(g)$. But then $H(g)\circ H(f)$ is the identity natural transformation from $H_d$ to itself, and $H(f)\circ H(g)$ is the identity natural transformation from $H_c$ to itself. We then have $H(f\circ g)=H(g)\circ H(f)=1_{H_d}=H(1_d)$ and $H(g\circ f)=H(f)\circ H(g)=1_{H_c}=H(1_c)$. By Yoneda's Lemma we must have $f\circ g=1_d$ and $g\circ f=1_c$. Thus $f$ and $g$ are mutual inverses and both are isomorphisms.