let $\psi: A \to B$ be an injective $R$ module homomorphism, and it is given that any $f: A \to M$ $\mathbb{Z}$ module homomorphism can be lifted to a $\mathbb{Z}$ module homomorphism $F: B \to M$ s.t $f=F \circ \psi$ $\DeclareMathOperator{\Hom}{Hom}$

To prove that any $f': A \to \Hom_{\mathbb{Z}}(R,M)$ arbitrary $R$ module homomorphism can be lifted to $F': B \to \Hom_{\mathbb{Z}}(R,M)$ $R$ module homomorphism with $f'=F' \circ \psi$

It is an exercise in Dummit Foote now I was trying to solve this(following hint in the book):

Consider $f(a)=f'(a)(1_R)\in M$ then as $f'(a) \in\Hom_{\mathbb{Z}}(R,M)$ we can see that $f: A \to M$ is a $\mathbb{Z}$ module homomorphism then it can be lifted to a $\mathbb{Z}$ module homomorphism $F: B \to M$ s.t $f=F \circ \psi$[ according to the hyp] this means

  1. $F(\psi(a))=f(a)=f'(a)(1_R)$

Now construct $F': B \to \Hom_{\mathbb{Z}}(R,M)$ s.t $F'(b)(r):=F(rb)$.

Claim that $F'$ is the lift of $f'$

Now there are two things to check

  • $F'$ is an $R$ module homomorphism i.e $F'(rb)=rF'(b)$

    Now $rF'(b)(s)=F'(b)(sr)$ for any $s \in R$[By the action of $R$ module on $Hom_{\mathbb{Z}}(R,M)$]

$\Rightarrow rF'(b)(s)=F(srb)=F'(rb)(s)\Rightarrow rF'(b)=F'(rb)$.

So $F'$ is an $R$ module homomorphism.

  • $f'=F' \circ \psi$.

Now here I am having problem. This is clear to me that $(F' \circ \psi(a))(1_R)=F'( \psi(a))(1_R)=F(1_R\psi(a))=F(\psi(a))=f'(a)(1_R)$[from eqn 1]

but I can't understand why $(F' \circ \psi(a))(r)=f'(a)(r)$ I was trying to think in this way that $f'(a)(r)=f'(a)(r1_R)$ but $f'(a)$ is a $\Bbb Z$ module homomorphism so I can't do anything from here. From L.H.S $(F' \circ \psi(a))(r)=F(r\psi(a))$. Now what?

I think I am missing some silly observation but I have to admit that I am lost.

Now one more non trivial question from my point of view that will there be any harm if we consider $f(a):=f'(a)(r_1)$ for some fixed $r_1 \in R$ instead of choosing $f(a)=f'(a)(1_R)$ and follow the same process from there? Please give some detailed explanation or a hint from where I can conclude.

  • 1
    $\begingroup$ $f'(ar)(1_R) = (f'(a) \cdot r)(1_R)= f'(a)(r)$. I explain this below. $\endgroup$ – Dean Young Apr 26 at 21:31

Here I answer why $(F' \circ \psi(a))(r) = f'(a)(r)$. It is (1) and (2) in the centerlined equality below. But before I begin, may I recommend that we start with the following setup instead:

Theorem 1: Let $R$ be a ring, and let $A$ and $B$ be right $R$-modules, and let $\psi : A \rightarrow B$ be an injective $R$-module morphism. Let $M$ be an abelian group, and regard $\text{Hom}_{\mathbb{Z}}( R, M)$ as being a right $R$-module where $(\phi \cdot r)(s) = \phi(rs)$. Suppose that, for every map of abelian groups $f : A \rightarrow M$, there is a map $g : B \rightarrow M$ of abelian groups such that $g \circ \psi = f$. Then, for each map of $R$-modules $f' : A \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$, there is a map $g' : B \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$ of $R$-modules such that $g' \circ \psi = f'$.

The difference is that $A$ and $B$ are right $R$-modules. This modification has the advantage of being consistent with the following heuristic:

Heuristic: In deciding whether a module should be a left, or right module, we have made the simplest and easiest choice when, for each equality involved, the variables occur in the same order in each term term. For instance, in choosing $\text{Hom}_{\mathbb{Z}}(R, M)$ to be a right $R$-module, $r$ always occurs before $s$ in $(\phi \cdot r)(s) = \phi(rs)$.

Proof of Theorem 1: take $f' : A \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$. $f'$ induces a $\mathbb{Z}$-linear $f : A \rightarrow M$ sending $a$ to $f'(a)(1_R)$ i.e $f(a)=f'(a)(1_R)$. $f$ induces a lift $F : B \rightarrow M$ by assumption. $F$ induces an $R$-linear map $F'$ sending $n$ to the map $R \rightarrow M$ sending $r$ to $F(nr)$. That is, $F'(n)(r) = F(nr)$. To see that $F'$ is a lift of $f'$, take $a \in A$ and $r \in R$. Then, $$(F' ( \psi(a)))(r) = F(\psi(a) r) = F(\psi(ar)) = f(ar) = f'(ar)(1_R) \stackrel{(1)}{=} (f'(a) \cdot r)(1_R) \stackrel{(2)}{=} f'(a)(r)$$ To see (1), note that $f' : A \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$ is an $R$-linear map of right $R$-modules. To see (2), note that the right $R$-module structure on $\text{Hom}_{\mathbb{Z}}(R, M)$ is given by $(\phi \cdot r)(s) = \phi(rs)$.

In this extra section, I explain how to go back and forth between right $R$-module maps $$ N \rightarrow \text{Hom}_{\mathbb{Z}} (R, M) $$ and $\mathbb{Z}$-module maps $$ N \rightarrow M$$

Take a ring $R$ Let $N$ be a right $R$-module. Let $M$ be a left $R$-module. $\text{Hom}_{\mathbb{Z}} (R, M)$ is a right $R$-module. We set $ \phi \cdot r : R \rightarrow M$ to be the map sending $s$ to $\phi(rs)$. Then $$(\phi \cdot (rs))(t) = \phi(rst) = (\phi \cdot r)(st) = ((\phi \cdot r )\cdot s) (t)$$

1) For the first direction, given a map $f : N \rightarrow \text{Hom}_{\mathbb{Z}} (R, M)$, we seek to define a map $g : N \rightarrow M$ of abelian groups. Set $g(n) = f(n)(1_R)$. Then $$g(n +m)= f(n+m)(1_R) = (f(n) + f(m))(1_R) = f(n)(1_R) + f(m)(1_R) = g(n) + g(m)$$

2) For the other direction, given a map $g : N \rightarrow M$ of $\mathbb{Z}$-modules, we seek to define a map $f : N \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$ of right $R$-modules. Set $f(n)(r) = g(nr)$. We check that $f : N \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$ is a map of right $R$-modules. Take $r \in R$ and $n \in N$. Take $s \in R$. Then $$f(nr)(s) = g(nrs) = f(n)(rs) = (f(n) \cdot r) (s)$$ So $f(nr) = f(n) \cdot r$. Note $f(n) \cdot r$ was defined above- we had to make the right choice as to whether $\text{Hom}_{\mathbb{Z}}(R, M)$ was a left or a right $R$-module.

Now to check that these operations are inverse. Take $f : N \rightarrow \text{Hom}_{\mathbb{Z}}(R, M)$ a map of right $R$-modules, and put $g : N \rightarrow M$ the map sending $n$ to $f(n)(1_R)$. Put $f_2$ the map sending $n$ to the map sending $r$ to $g(nr)$. Then $f_2 (n) = f(n)$ for each $n$. Indeed, for each $n \in N$ and each $r \in R$, $$ f_2(n)(r) = g(nr) = f(nr)(1_R) = (f(n) \cdot r) (1_R) = f(n)(r 1_R) = f(n)(r)$$ notice how in all these terms, $n$ always occurs before $r$. That is the heuristic that we have followed to make sure all the parities match up. So $f = f_2$.

Next, take a map of abelian groups $g : N \rightarrow M$. Set $f$ to be the map sending $n$ to the map $f(n) : R \rightarrow M$ of abelian groups sending $r$ to $g(nr)$. Then set $g_2 (n) : N \rightarrow M$ to be the map of abelian groups sending $n$ to $f(n)(1_R)$. Then $$g_2 (n) = f(n)(1_R) = g(n1_R) = g(n)$$

This establishes the desired correspondence.

  • $\begingroup$ I think it would $g:B \to M$ in the highlighted portion. $\endgroup$ – Shadow Apr 26 at 23:22
  • $\begingroup$ There were some typos e.g $($ was missing and $A$ was written instead of $B$ in the first highlighted section and also I suggested a different notation $f'$ and $g'$ in that part. It is a detailed nice answer. Thanks a lot. $\endgroup$ – Shadow Apr 26 at 23:37
  • $\begingroup$ You're welcome, Shadow. Did the errors get fixed? $\endgroup$ – Dean Young Apr 27 at 3:01
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    $\begingroup$ Yes, thanks a lot. :) $\endgroup$ – Shadow Apr 27 at 23:00

This is really just the tensor-hom adjunction. We have $$ \mathrm{Hom}_R(A,\mathrm{Hom}_{\mathbb Z}(R,M)) \cong \mathrm{Hom}_{\mathbb Z}(R\otimes_RA,M) \cong \mathrm{Hom}_{\mathbb Z}(A,M). $$ Explicitly, we can identify $R$-linear maps $A\to\mathrm{Hom}_{\mathbb Z}(R,M)$ with $\mathbb Z$-linear maps $A\to M$, by sending $f'$ to $f$ such that $f(a)=f'(a)(1_R)$, and sending $f$ to $f'$ such that $f'(a)(r)=f(ra)$.

Note that the $R$-action on $\mathrm{Hom}_{\mathbb Z}(R,M)$ is given by $r\theta\colon s\mapsto \theta(sr)$, so given $f$, the map $f'$ such that $f'(a)(r)=f(ra)$ really is $R$-linear: $$ f'(ra)(s) = f(sra) = f'(a)(sr) = (rf'(a))(s). $$

Following this through, we start with an $R$-linear map $f'\colon A\to\mathrm{Hom}_{\mathbb Z}(R,M)$, obtain a $\mathbb Z$-linear map $f\colon A\to M$, lift to a $\mathbb Z$-linear map $F\colon B\to M$ such that $F\psi=f$, and obtain the corresponding $R$-linear map $F'\colon B\to\mathrm{Hom}_{\mathbb Z}(R,M)$.

We just need to check that $F'\psi=f'$. By construction we have $$ F'(\psi(a))(r) = F(r\psi(a)) = F(\psi(ra)) = f(ra) = f'(a)(r). $$ This holds for all $r\in R$, so $F'\psi(a)=f'(a)$. This holds for all $a\in A$, so $F'\psi=f'$.

  • $\begingroup$ From $f(a)=f'(a)(1_R)$ why is $f(ra)=f'(a)(r)$? $\endgroup$ – Shadow Apr 26 at 18:37
  • $\begingroup$ In your notation, $A$ and $B$ are left $R$-modules. Therefore $\text{Hom}_{\mathbb{Z}}(R, M)$ must be a left $R$-module for this to make sense. Therefore $R$ in $\text{Hom}_{\mathbb{Z}}(R, M)$ must be a right $R$-module for this to make sense. Therefore the correct multiplication of $R$ on $\text{Hom}_{\mathbb{Z}}(R, M)$ sends $(r, \phi )$ to the map $R \rightarrow M$ sending $s$ to $\phi(sr)$, not $\phi(rs)$. Of course, it's not a problem if we take $R$ to be commutative here :) $\endgroup$ – Dean Young Apr 26 at 20:36
  • $\begingroup$ In the original post we are considering left $R$-modules, so I continued with this. I then wrote explicitly that the left $R$-action on $\mathrm{Hom}_{\mathbb Z}(R,M)$ is given by $r\theta\colon s\mapsto \theta(sr)$. $\endgroup$ – Andrew Hubery Apr 26 at 23:18
  • $\begingroup$ My bad- this should work. So this suggests that there are two forms of the hom-tensor adjunction- one where $\text{Hom}_R (S, -) : R \text{-mod} \rightarrow S \text{-mod}$ is hom as right $R$-modules, and another where it is hom as left $R$-modules. $\endgroup$ – Dean Young Apr 27 at 2:50

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