# Any $f': A \to \mathrm{Hom}_{\mathbb{Z}}(R,M)$ can be lifted to $F': B \to \mathrm{Hom}_{\mathbb{Z}}(R,M)$

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.

• $f'(ar)(1_R) = (f'(a) \cdot r)(1_R)= f'(a)(r)$. I explain this below. – 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.

• I think it would $g:B \to M$ in the highlighted portion. – Shadow Apr 26 at 23:22
• 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. – Shadow Apr 26 at 23:37
• You're welcome, Shadow. Did the errors get fixed? – Dean Young Apr 27 at 3:01
• Yes, thanks a lot. :) – 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 $$F'\psi=f'$$.

• From $f(a)=f'(a)(1_R)$ why is $f(ra)=f'(a)(r)$? – Shadow Apr 26 at 18:37
• 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 :) – Dean Young Apr 26 at 20:36
• 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 $r\theta\colon s\mapsto \theta(sr)$. – Andrew Hubery Apr 26 at 23:18
• 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. – Dean Young Apr 27 at 2:50