The Contravariant Functor Hom_R( _ , X)

[Math Amateur]

I am reading Paul E. Bland's book, "Rings and Their Modules".

I am trying to understand Section 3.1 on categories and need help in understanding the contravariant functor \(\displaystyle \text{Hom}_R(\_, X)\) as described in Bland, Example 13 in Ch. 3: Categories (page 76).

Example 13 in Ch. 3 reads as follows:
https://www.physicsforums.com/attachments/3635Now, in the functor \(\displaystyle \text{Hom}_R(\_, X) \ : \ Mod_R \longrightarrow Ab\), \(\displaystyle X\) is a fixed \(\displaystyle R\)-module.

The functor \(\displaystyle \text{Hom}_R(\_, X)\) assigns each object \(\displaystyle M\) in the category \(\displaystyle \text{Mod}_R\) to an object \(\displaystyle \text{Hom}_R (M,X) \) in Ab.

(Note that we know that \(\displaystyle \text{Hom}_R(A,B) \) where \(\displaystyle A\) and \(\displaystyle B\) are \(\displaystyle R\)-modules is an abelian group!)
Now \(\displaystyle \text{Hom}_R(\_, X)\) must also assign each morphism \(\displaystyle f\) in \(\displaystyle \text{Mod}_R\) to a morphism \(\displaystyle f^*\) in Ab. So Bland defines the following assignment of \(\displaystyle f\) to \(\displaystyle f^*\):\(\displaystyle \text{Hom}_R(\_, X) (f) = \text{Hom}_R(f, X) = f^* \)

where \(\displaystyle f^* \ : \ \text{Hom}_R(N, X) \longrightarrow \text{Hom}_R(M, X)\)

is given by \(\displaystyle f^*(h) = hf\)... ... BUT ... ... what exactly is \(\displaystyle h\) ... clearly I need to understand the nature of h to understand \(\displaystyle f^*\) ... ... can someone please help me with this matter?
Just to show my own thinking and thus specifically why I have a problem ... see the following ...It seems that ... ... \(\displaystyle f \ : \ \text{Mod}_R \longrightarrow \text{Mod}_R \ \ \text{ where } \ \ f \ : \ M \longrightarrow N\).Now we need \(\displaystyle f\) to map to \(\displaystyle f^*\) where \(\displaystyle f^* \ : \ \text{Ab} \longrightarrow \text{Ab} \) ... ... So an \(\displaystyle f^*\) defined by

\(\displaystyle f^* \ : \ \text{Hom}_R(N, X) \longrightarrow \text{Hom}_R(M, X)\)

will do ... ... since \(\displaystyle \text{Hom}_R(N, X)\) and \(\displaystyle \text{Hom}_R(M, X)\) are abelian groups.Now since \(\displaystyle f^*\) is defined as \(\displaystyle f(h) = hf\) we must have \(\displaystyle h \in \text{Hom}_R(N, X)\) and so \(\displaystyle h\) is of the form \(\displaystyle h \ : \ N \longrightarrow X\).


So we have

\(\displaystyle f \ : \ M \longrightarrow N\) and \(\displaystyle h \ : \ N \longrightarrow X\)

so then ...

\(\displaystyle f^* = hf \ : \ M \longrightarrow X
\)... ... BUT ... problem ... ... \(\displaystyle f^*\) should be a mapping between \(\displaystyle \text{Hom}_R(N, X)\) and \(\displaystyle \text{Hom}_R(M, X)\) ... ... and not a mapping between two \(\displaystyle R\)-modules, \(\displaystyle M\) and \(\displaystyle X\).

Can someone please clarify this for me?

Further, can someone criticize my analysis/thinking above?

Help will be appreciated ...

Peter
***NOTE***

I think it may be helpful for MHB members reading this post to be able to see Bland's definition of a functor.

Bland's definition of a functor, therefore, is provided below:View attachment 3636
View attachment 3637

 

[Euge]

Hi Peter,

I think the issue you're having lies in the notation $f^*$. Note that $f^*$ maps an object $h \in \text{Hom}_R(N,X)$ to the object $hf \in \text{Hom}_R(M, X)$ (think of pre-composition). So the assignment $f^* : h \to hf$ is a function from $\text{Hom}_R(N,X)$ to $\text{Hom}_R(M,X)$. To show that $f^*$ is a morphism in Ab, you need to verify that $f^*$ is a homomorphism of abelian groups. The Hom-sets involved are abelian groups under (pointwise) addition of functions, so you need to show that $f^*(h + h') = f^*(h) + f^*(h')$ for all $h, h' \in \text{Hom}_R(N,X)$.

 

[Math Amateur]

Hi Peter,

I think the issue you're having lies in the notation $f^*$. Note that $f^*$ maps an object $h \in \text{Hom}_R(N,X)$ to the object $hf \in \text{Hom}_R(M, X)$ (think of pre-composition). So the assignment $f^* : h \to hf$ is a function from $\text{Hom}_R(N,X)$ to $\text{Hom}_R(M,X)$. To show that $f^*$ is a morphism in Ab, you need to verify that $f^*$ is a homomorphism of abelian groups. The Hom-sets involved are abelian groups under (pointwise) addition of functions, so you need to show that $f^*(h + h') = f^*(h) + f^*(h')$ for all $h, h' \in \text{Hom}_R(N,X)$.

Thanks so much for that help Euge ...

I am still reflecting on what you have said ... but I was quite perplexed and now things are becoming clearer ...

Thanks again ...

Peter