Homework 5

We have seen that $\mathbf{C}{\otimes }_{\mathbf{C}}\mathbf{C}\simeq \mathbf{C}$ as $\mathbf{C}$-modules, and hence also as $\mathbf{R}$-modules. We also know that $\mathbf{C}\simeq {\mathbf{R}}^2\simeq \mathbf{R}\oplus \mathbf{R}$ as $\mathbf{R}$-modules (from linear algebra, say).

On the other hand, since tensor product commutes with direct sums, we have $\mathbf{R}$-module isomorphisms

$\mathbf{C}{\otimes }_{\mathbf{R}}\mathbf{C}\simeq \mathbf{C}{\otimes }_{\mathbf{R}}(\mathbf{R}\oplus \mathbf{R})\simeq (\mathbf{C}{\otimes }_{\mathbf{R}}\mathbf{R})\oplus (\mathbf{C}{\otimes }_{\mathbf{R}}\mathbf{R})\simeq \mathbf{C}\oplus \mathbf{C}\simeq (\mathbf{R}\oplus \mathbf{R})\oplus (\mathbf{R}\oplus \mathbf{R})\simeq {\mathbf{R}}^4.$

In the language of linear algebra, $\mathbf{C}{\otimes }_{\mathbf{R}}\mathbf{C}$ is isomorphic as a real vector space to ${\mathbf{R}}^4$, while $\mathbf{C}{\otimes }_{\mathbf{C}}\mathbf{C}$ is isomorphic as a real vector space to ${\mathbf{R}}^2$. Since isomorphisms of vector spaces preserve dimension, it follows that $\mathbf{C}{\otimes }_{\mathbf{R}}\mathbf{C}$ and $\mathbf{C}{\otimes }_{\mathbf{C}}\mathbf{C}$ cannot be isomorphic as $\mathbf{R}$-modules.

First note that the zero elements in $Q$ and $D$ coincide, and $0_Q\otimes m=(1_Q\cdot 0_D)\otimes m=1_Q\otimes (0_D\cdot m)=1_Q\otimes 0_m$, so the result holds for simple tensors of the form $0_Q\otimes m$.

Now suppose $q\otimes m$ is a simple tensor with $q\in Q$ nonzero and $m\in M$. We can write $q=\frac{a}{d}$ for some $a,b\in D$ with $b$ nonzero, and we then have

$q\otimes m=\frac{a}{b}\otimes m=(\frac{1}{b}\cdot a)\otimes m=\frac{1}{b}\otimes am,$

so the result holds for $q\otimes m$.

Finally, suppose we take a general element of $Q{\otimes }_{D}M$, represented as a finite sum of the form ${\displaystyle \sum _{i=1}^k{q}_i\otimes m_i}$. Using the above argument, we can write each simple tensor in this sum in the form $q_i\otimes m_i=\frac{1}{b_i}\otimes m_i^{\prime }$ for some $b_i\in D$ nonzero and $m_i^{\prime }\in M$. Let $d$ be a least common multiple of the $b_i$ in $D$, so that for each $i$ we can write $\frac{1}{b_i}=\frac{c_i}{d}$ for some $c_i\in D$. We can then rewrite each simple tensor as

$\frac{1}{b_i}\otimes m_i^{\prime }=\frac{c_i}{d}\otimes m_i^{\prime }=\frac{1}{d}\otimes c_i{m}_i^{\prime }.$

Using linearity, we then have

$\sum _{i=1}^n(q_i\otimes m_i)=\sum _{i=1}^n(\frac{1}{b_i}\otimes m_i^{\prime })=\sum _{i=1}^n(\frac{1}{d}\otimes c_i{m}_i^{\prime })=\frac{1}{d}\otimes \sum _{i=1}^n{c}_i{m}_i^{\prime }=\frac{1}{d}\otimes m,$

as desired.

We have seen that a basis for ${\mathbf{R}}^2\otimes {\mathbf{R}}^2$ as an $\mathbf{R}$-module is $\{e_1\otimes e_1,e_1\otimes e_2,e_2\otimes e_1,e_2\otimes e_2\}$. In particular, these four tensors are linearly independent over $\mathbf{R}$. Now suppose, towards a contradiction, that $e_1\otimes e_2+e_2\otimes e_1=v\otimes w$ for some $v,w\in {\mathbf{R}}^2$. Let's expand the right-hand side in terms of our basis. We have $v=a_1{e}_1+a_2{e}_2$ and $w=b_1{e}_1+b_2{e}_2$ for some unique real numbers $a_1,a_2,b_1,b_2\in \mathbf{R}$. We then have

$\begin{array}{rl}v\otimes w& =(a_1{e}_1+a_2{e}_2)\otimes (b_1{e}_1+b_2{e}_2)\\ & =a_1{b}_1(e_1\otimes e_2)+a_1{b}_2(e_1\otimes e_2)+a_2{b}_1(e_2\otimes e_1)+a_2{b}_2(e_2\otimes e_2).\end{array}$

Since the four tensors are a basis for our vector space, the equality $e_1\otimes e_2+e_2\otimes e_1=v\otimes w$ forces the following four equalities:

1. $a_1{b}_1=0$
2. $a_2{b}_1=1$
3. $a_1{b}_2=1$
4. $a_2{b}_2=0$

Equations (2) and (3) imply that all four numbers $a_1,a_2,b_1,b_2$ are nonzero (as $\mathbf{R}$ is a domain), but then equations (1) and (4) are impossible.

Thus, the tensor $e_1\otimes e_2+e_2\otimes e_1$ cannot be written as a simple tensor.

Consider the extension of scalars from $\mathbf{Z}$ to $\mathbf{Q}$ of the direct product of the $\mathbf{Z}$-modules $M_i={\mathbf{Z}}_2$ for $i=1,2,\dots$ .

On the one hand, since each $M_i$ is a finite abelian group, we know that $\mathbf{Q}{\otimes }_{\mathbf{Z}}{M}_i=0$ for every $i$. It follows that ${\displaystyle \prod _{i=1}^{\mathrm{\infty }}\mathbf{Q}{\otimes }_{\mathbf{Z}}{M}_i=0}$.

On the other hand, we claim that ${\displaystyle \mathbf{Q}{\otimes }_{\mathbf{Z}}\prod _{i=1}^{\mathrm{\infty }}{M}_i}$ is nonzero. To see this, first note that $\mathbf{Q}$ is the field of fractions of the integral domain $\mathbf{Z}$, so by our general properties for any $\mathbf{Z}$-module $M$ the $\mathbf{Q}$-module $\mathbf{Q}{\otimes }_{\mathbf{Z}}M$ is zero exactly when $M$ is torsion (as a $\mathbf{Z}$-module). However, in our case the $\mathbf{Z}$-module ${\displaystyle M=\prod _{i=1}^{\mathrm{\infty }}{M}_i}$ has many elements that are not torsion; e.g., the element $m=(1,1,1,\dots )$ is not torsion. (For any integer $k\in \mathbf{Z}$, simply note that for all $i$ with $2^i>|k|$ the element $k\cdot m=(k,k,k,\dots )$ is nonzero in the $i$th coordinate.) We thus have that $\mathbf{Q}{\otimes }_{\mathbf{Z}}M$ is nonzero, and hence is not isomorphic to the zero module ${\displaystyle \prod _{i=1}^{\mathrm{\infty }}\mathbf{Q}{\otimes }_{\mathbf{Z}}{M}_i}$.

1. First observe that $i_1(r_1+r_2)=(r_1+r_2)\otimes 1_S=r_1\otimes 1_S+r_2\otimes 1_S=i_1(r)+i_2(r),$
   and also
   $i_2(r_1{r}_2)=r_1{r}_2\otimes 1_S=r_1{r}_2\otimes 1_S\cdot 1_S=(r_1\otimes 1_S)\cdot (r_2\otimes 1_S)=i_1(r_1)\cdot i_1(r_2).$
   This prove $i_1$ is a ring morphism. The same argument, mutatis mutandis, proves that $i_2$ is a ring morphism.

2. We need to prove that for every pair of ring morphisms $f:R\to T$, $g:S\to T$ there is a unique ring morphism $h:R\otimes S\to T$ through which $f$ and $g$ factor. So, suppose $f:R\to T$, $g:S\to T$ is a pair of ring morphisms. Define a set map $\varphi :R\times S\to T$ by $\varphi (r,s)=f(r)g(s)$. Observe that this is a bilinear map:

   $\begin{array}{rl}\varphi ((r_1,s)+(r_2,s))& =\varphi (r_1+r_2,s)=f(r_1+r_2)g(s)\\ & =(f(r_1)+f(r_2))g(s)\\ & =f(r_1)g(s)+f(r_2)g(s)\\ & =\varphi (r_1,s)+\varphi (r_2,s)\end{array}$

   and similarly $\varphi ((r,s_1)+(r,s_2))=\varphi (r,s_1)+\varphi (r,s_2).$

   By the universal property of the tensor product, we then have a corresponding ring morphism $h:R\otimes S\to T$ defined on simple tensors by $h(r\otimes s)=f(r)g(s)$. Then note that

   $(h\circ i_1)(r)=h(r\otimes 1_S)=f(r)g(1_S)=f(r)1_T=f(r),$

   so we indeed have $f=h\circ i_1$. We similarly have $g=h\circ i_2$, since

   $(h\circ i_2)(s)=h(1_R\otimes s)=f(1_R)g(s)=1_{T}g(s)=g(s).$

   So we've shown $f$ and $g$ factor through $h$.

   As for uniqueness, suppose $h^{\prime }:R\otimes S\to T$ is another ring morphism such that $f$ and $g$ factor through $h^{\prime }$. Then observe that

   $\begin{array}{rl}{h}^{\prime }(r\otimes s)& =h^{\prime }((r\otimes 1_S)\cdot (1_R\otimes s))\\ & =h^{\prime }(r\otimes 1_S)\cdot h^{\prime }(1_R\otimes s)\\ & =h^{\prime }(i_1(r))\cdot h^{\prime }(i_2(s))\\ & =f(r)g(s)\\ & =h(r\otimes s).\end{array}$

   This proves uniqueness of $h$, which establishes that $R\otimes S$ (together with the ring morphisms $i_1,i_2$) satisfy the universal property of the coproduct of $R$ and $S$ (in the category of commutative rings).

For those who might be curious, here we'll present a full solution covering (nearly) every detail one should officially check for this problem.

The abelian group structure

We first show ${\mathrm{Hom}}_S(M,N)$ is an abelian group. The addition in ${\mathrm{Hom}}_S(M,N)$ is defined by addition of outputs; i.e., for every pair of right $S$-module morphisms $f,g:M\to N$, we define $f+g:M\to N$ by $(f+g)(m)=f(m)+g(m)$. Of course, we need to verify this operation is well defined; i.e., that $f+g$ is still a right $S$-module morphism. It is additive because $f$ and $g$ are additive.

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

It is compatible with the right $S$-action because $f$ and $g$ are compatible with the right $S$-action:

$\begin{array}{rl}(f+g)(m)\cdot s& =(f(m)+g(m))\cdot s\\ & =f(m)\cdot s+g(m)\cdot s\\ & =f(ms)+g(ms)\\ & =(f+g)(ms).\end{array}$

So we have a well-defined binary operation on ${\mathrm{Hom}}_S(M,N)$. It is evidently commutative and associative, since addition in $N$ is commutative and associative. The zero map from $M$ to $N$ is a right $S$-module morphism, and it plays the role of the additive identity under this operation. Additive inverses are given via additive inverses of outputs; i.e., $(-f)(n)=-f(n)$. It is straightforward to verify $-f$ is indeed another right $S$-module morphism; let's skip checking that, at least.

The left $R^{\prime }$-module structure

We've thus established that ${\mathrm{Hom}}_S(M,N)$ has the structure of an abelian group. We next consider the left and right actions.

We define the left $R^{\prime }$-action using the left $R^{\prime }$-action on $N$, namely by defining $r^{\prime }\cdot f$ as the function given by $(r^{\prime }\cdot f)(m)=r^{\prime }\cdot f(m)$. Once again, we need to verify this is well defined; i.e., the map $r^{\prime }\cdot f$ is another right $S$-module morphism. The map $r^{\prime }\cdot f$ is certainly additive:

$\begin{array}{rl}(r^{\prime }\cdot f)(m_1+m_2)& =r^{\prime }\cdot f(m_1+m_2)\\ & =r^{\prime }\cdot (f(m_1)+f(m_2))\\ & =r^{\prime }\cdot f(m_1)+r^{\prime }\cdot f(m_2)\\ & =(r^{\prime }\cdot f)(m_1)+(r^{\prime }\cdot f)(m_2).\end{array}$

It is still compatible with the right $S$-action:

$\begin{array}{rl}(r^{\prime }\cdot f)(m)\cdot s& =r^{\prime }\cdot f(m)\cdot s\\ & =r^{\prime }\cdot f(ms)\\ & =(r^{\prime }\cdot f)(ms).\end{array}$

So, $r^{\prime }\cdot f$ is indeed another right $S$-module morphism from $M$ to $N$.

To verify this defines a left action of $R^{\prime }$ on ${\mathrm{Hom}}_S(M,N)$, observe that we certainly have $1_{R^{\prime }}\cdot f=f$, since for all $m\in M$ we have

$(1_{R^{\prime }}\cdot f)(m)=1_{R^{\prime }}\cdot f(m)=f(m).$

Similarly, we have $(r_1^{\prime }+r_2^{\prime })\cdot f=r_1^{\prime }\cdot f+r_2^{\prime }\cdot f$, as

$\begin{array}{rl}((r_1^{\prime }+r_2^{\prime })\cdot f)(m)& =(r_1^{\prime }+r_2^{\prime })\cdot f(m)\\ & =r_1^{\prime }\cdot f(m)+r_2^{\prime }\cdot f(m)\\ & =(r_1^{\prime }\cdot f)(m)+(r_2^{\prime }\cdot f)(m).\end{array}$

It is also compatible with our additive operation in ${\mathrm{Hom}}_S(M,N)$:

$\begin{array}{rl}(r^{\prime }\cdot (f_1+f_2))(m)& =r^{\prime }\cdot (f_1+f_2)(m)\\ & =r^{\prime }\cdot (f_1(m)+f_2(m))\\ & =r^{\prime }\cdot f_1(m)+r^{\prime }\cdot f_2(m)\\ & =(r^{\prime }\cdot f_1)(m)+(r^{\prime }\cdot f_2)(m)\\ & =(r^{\prime }\cdot f_1+r^{\prime }\cdot f_2)(m),\end{array}$

thus $r^{\prime }\cdot (f_1+f_2)=r^{\prime }\cdot f_1+r^{\prime }\cdot f_2$.

Lastly, we should verify that $r_1^{\prime }\cdot (r_2^{\prime }\cdot f)=(r_1^{\prime }{r}_2^{\prime })\cdot f$ for all $r_1^{\prime },r_2^{\prime }\in R^{\prime }$. To see this is true, observe that for every $m\in M$ we have

$(r_1^{\prime }\cdot (r_2^{\prime }\cdot f))(m)=r_1^{\prime }\cdot ((r_2^{\prime }\cdot f)(m))=r_1^{\prime }\cdot (r_2^{\prime }\cdot f(m))=(r_1^{\prime }{r}_2^{\prime })\cdot f(m)=(r_1^{\prime }{r}_2^{\prime }\cdot f)(m),$

where the second-to-last equality holds by the fact that we have a left $R^{\prime }$-action on $N$.

The right $R$-module structure

We define the right $R$-module structure on ${\mathrm{Hom}}_S(M,N)$ using the left $R$-action on $M$, namely by defining $f\cdot r$ as the function given by $(f\cdot r)(m)=f(rm)$. We now repeat the same analysis as above to verify this is a well-defined right $R$-action on ${\mathrm{Hom}}_S(M,N)$.

The map $f\cdot r$ is additive:

$\begin{array}{rl}(f\cdot r)(m_1+m_2)& =f(r(m_1+m_2))\\ & =f(r{m}_1+r{m}_2))\\ & =f(r{m}_1)+f(r{m}_2)\\ & =(f\cdot r)(m_1)+(f\cdot r)(m_2).\end{array}$

It is still compatible with the right $S$-action:

$\begin{array}{rl}(f\cdot r)(m)\cdot s& =f(rm)\cdot s\\ & =f(rms)\\ & =(f\cdot r)(ms).\end{array}$

So, $f\cdot r$ is indeed another right $S$-module morphism from $M$ to $N$.

To verify this defines a right action of $R$ on ${\mathrm{Hom}}_S(M,N)$, observe that we certainly have $f\cdot 1_R=f$, since for all $m\in M$ we have

$(f\cdot 1_R)(m)=f(1_R\cdot m)=f(m),$

and similarly $f\cdot (r_1+r_2)=f\cdot r_1+f\cdot r_2$, as

$\begin{array}{rl}(f\cdot (r_1+r_2))(m)& =f((r_1+r_2)m)\\ & =f(r_{1}m+r_{2}m)\\ & =f(r_{1}m)+f(r_{2}m)\\ & =(f\cdot r_1)(m)+(f\cdot r_2)(m_2)\\ & =(f\cdot r_1+f\cdot r_2)(m).\end{array}$

It is also compatible with our additive operation in ${\mathrm{Hom}}_S(M,N)$:

$\begin{array}{rl}((f_1+f_2)\cdot r)(m)& =(f_1+f_2)(rm)\\ & =f_1(rm)+f_2(rm)\\ & =(f_1\cdot r)(m)+(f_2\cdot r)(m)\\ & =(f_1\cdot r+f_2\cdot r)(m),\end{array}$

thus $(f_1+f_2)\cdot r=f_1\cdot r+f_2\cdot r$.

Finally, we should verify that $(f\cdot r_1)\cdot r_2=f\cdot (r_1{r}_2)$ for all $r_1,r_2\in r$. To see this, observe that for every $m\in M$ we have

$((f\cdot r_1)\cdot r_2)(m)=(f\cdot r_1)(r_{2}m)=f(r_1(r_{2}m))=f((r_1{r}_2)m)=(f\cdot (r_1{r}_2))(m),$

where the second-to-last equality holds by the fact that we have a left $R$-action on $M$.

The bimodule condition

The last property we need to verify is that our two actions on ${\mathrm{Hom}}_S(M,N)$ are compatible. To that end, suppose $r^{\prime }\in R$, $r\in R$, and $f\in {\mathrm{Hom}}_S(M,N)$. For every $m\in M$ we have

$(r^{\prime }\cdot (f\cdot r))(m)=r^{\prime }\cdot (f\cdot r)(m)=r^{\prime }\cdot f(rm),$

while

$((r^{\prime }\cdot f)\cdot r)(m)=(r^{\prime }\cdot f)(rm)=r^{\prime }\cdot f(rm).$

Thus, we do indeed have $r^{\prime }\cdot (f\cdot r)=(r^{\prime }\cdot f)\cdot r$.

We've at last established ${\mathrm{Hom}}_S(M,N)$ does indeed have the structure of an $(R^{\prime },R)$-bimodule.

1. Suppose $f:P_1\to P_2$ is an $(R,T)$-bimodule morphism. For each $(R,T)$-bimodule morphism $g:M{\otimes }_{S}N\to P_1$, composition with $f$ yields an $(R,T)$-bimodule morphism $f\circ g:M{\otimes }_{S}N\to P_2$. This defines a set map from $F(P_1)={\mathrm{Hom}}_{(R,T)}(M{\otimes }_{S}N,P_1)$ to $F(P_2)={\mathrm{Hom}}_{(R,T)}(M{\otimes }_{S}N,P_2).$

   Optional functoriality check

   We can verify this map on arrows (together with the given map on objects) does indeed define a functor $F:(R,T)-\mathbf{\text{Mod}}\to \mathbf{\text{Set}}$. First note that for each identity arrow $1_P:P\to P$ we have $F(1_P)=1_P\circ -$, which sends each $(R,T)$-morphism $g:M{\otimes }_{S}N\to P$ to $1_P\circ g=g$; i.e., is the identity map on the set ${\mathrm{Hom}}_{(R,T)}(M{\otimes }_{S}N,P)=F(P)$.

   Also, for each pair of composable $(R,T)$-bimodule morphisms $f_1:P_1\to P_2$ and $f_2:P_2\to P_3$, for every $(R,T)$-module morphism $g:M{\otimes }_{S}N:\to P$ we have $F(f_2\circ f_1)(g)=(f_2\circ f_1)\circ g$ and $(F(f_2)\circ F(f_1))(g)=f_2\circ (f_1\circ g)$. By associativity of composition in $(R,T)-\mathbf{\text{Mod}}$ these compositions are always equal, hence $F(f_2\circ f_1)=F(f_2)\circ F(f_2)$. Thus, $F$ does indeed define a functor.

   ---

   We next consider the functor $G$. Suppose $f:P_1\to P_2$ is an $(R,T)$-bimodule morphism. For each $(R,S)$-bimodule morphism $g:M\to {\mathrm{Hom}}_T(N,P_1)$, let's introduce the temporary notation $g_m$ for the right $T$-module morphism $g(m):N\to P_1$. In particular, note that composition with $f$ then yields a right $T$-module morphism $f\circ g_m:N\to P_2$. In other words, $f\circ g_m\in {\mathrm{Hom}}_T(N,P_2).$ It therefore makes sense to define a set map $M\to {\mathrm{Hom}}_T(N,P_2)$ by $m\mapsto f\circ g_m$. Again, in order to avoid getting overwhelmed by parentheses, let's denote this map by $(Gf{)}_g$. Of course, we need to verify that $(Gf{)}_g$ is actually an $(R,S)$-bimodule morphism. We first note that $g$ is a module morphism and hence additive:

   $g_{m_1+m_2}=g(m_1+m_2)=g(m_1)+g(m_2)=g_{m_1}+g_{m_2}$

   Then we can observe that

   $\begin{array}{rl}(Gf{)}_g(m_1+m_2)& =f\circ g_{m_1+m_2}\\ & =f\circ (g_{m_1}+g_{m_2})\\ & =f\circ g_{m_1}+f\circ g_{m_2}\\ & =(Gf{)}_g(m_1)+(Gf{)}_g(m_2).\end{array}$

   We next verify $(Gf{)}_g$ is compatible with the left $R$-action. Observe that since both $f$ and $g$ are compatible with the left $R$-action we have

   $\begin{array}{rl}(Gf{)}_g(rm)& =f\circ g_{rm}\\ & =f\circ (g(rm))\\ & =f\circ (r\cdot g(m))\\ & =f\circ (r\cdot g_m)\\ & =r\cdot (f\circ g_m)\\ & =r\cdot (Gf{)}_g(m).\end{array}$

   It is also compatible with the right $S$-action, once we recall that the right $S$ action on ${\mathrm{Hom}}_T(N,P)$ is by left multiplication on the inputs:

   $\begin{array}{rl}(Gf{)}_g(ms)& =f\circ g_{ms}\\ & =f\circ (g(ms))\\ & =f\circ (g(m)\cdot s)\\ & =f\circ (g_m\cdot s)\\ & =f\circ (g_m(s\cdot -))\\ & =(f\circ g_m)(s\cdot -)\\ & =(f\circ g_m)\cdot s\\ & =(Gf{)}_g(m)\cdot s.\end{array}$

   So, we've verified $(Gf{)}_g$ is indeed an $(R,S)$-bimodule morphism and hence we have defined a set map

   $G(f):{\mathrm{Hom}}_{(R,S)}(M,{\mathrm{Hom}}_T(N,P_1))\to {\mathrm{Hom}}_{(R,S)}(M,{\mathrm{Hom}}_T(N,P_2))$

   that given by $g\mapsto (Gf{)}_g$.

   Optional Functoriality Check

   Lastly, we should verify this map on arrows (together with the given map on objects) does indeed define a functor $G:(R,T)-\mathbf{\text{Mod}}\to \mathbf{\text{Set}}$. First note that for each identity arrow $1_P:P\to P$ we have $G(1_P)$ is the set map $g\mapsto (G{1}_P{)}_g$. But for every $m\in M$ we have $(G{1}_P{)}_g(m)=1_P\circ g_m=g_m=g(m)$, which is the identity on $G(P)$.

   Also, for each pair of composable $(R,T)$-bimodule morphisms $f_1:P_1\to P_2$ and $f_2:P_2\to P_3$, for every $(R,S)$-module morphism $g:M\to {\mathrm{Hom}}_T(N,P_1)$, and for every $m\in M$ we have

   $G(f_2\circ f_1{)}_g(m)=(f_2\circ f_1)\circ g_m=f_2\circ (f_1\circ g_m)$

   and

   $(G(f_2)\circ G(f_1){)}_g(m)=G(f_2{)}_{G(f_1{)}_g}(m)=f_2\circ (G(f_1{)}_g(m))=f_2\circ (f_1\circ g_m)).$

   By associativity of composition we therefore have $G(f_2\circ f_1{)}_g=(G(f_2)\circ G(f_1){)}_g$, and as $g$ was arbitrary, this prove $G(f_2\circ f_1)=G(f_2)\circ G(f_1)$.

2. Suppose $f:P_1\to P_2$ is an $(R,T)$-bimodule morphism. We need to verify that the diagram below is commutative:

   Take any element in the top-left corner; i.e., a $(R,T)$-bimodule morphism $g:M{\otimes }_{S}N\to P_1$. The left vertical map sends $g$ to the $(R,T)$-bimodule morphism $f\circ g:M{\otimes }_{S}N\to P_2$. The bottom horizontal map then sends $f\circ g$ to the $(R,S)$-bimodule morphism ${\tau }_{P_2}(f\circ g):M\to {\mathrm{Hom}}_T(N,P_2)$ that assigns to each $m\in M$ the right $T$-module morphism $n\mapsto (f\circ g)(m\otimes n)$.

   On the other hand, the top horizontal map sends $g$ to the $(R,S)$-bimodule morphism ${\tau }_{P_1}(g):M\to {\mathrm{Hom}}_T(N,P_1)$ that assigns to each $m\in M$ the right $T$-module morphism $n\mapsto g(m\otimes n)$. The right vertical map then sends ${\tau }_{P_1}(g)$ to the $(R,S)$-bimodule morphism $(Gf{)}_{{\tau }_{P_1}(g)}:M\to {\mathrm{Hom}}_T(N,P_2)$ that assigns to each $m\in M$ the right $T$-module morphism $f\circ {\tau }_{P_1}(g{)}_m$, which is the map $n\mapsto f(g(m\otimes n))$.

   Thus, the diagram is indeed commutative.