MHB 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 $$\text{Hom}_R(\_, X)$$ 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 $$\text{Hom}_R(\_, X) \ : \ Mod_R \longrightarrow Ab$$, $$X$$ is a fixed $$R$$-module.

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

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

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

is given by $$f^*(h) = hf$$... ... BUT ... ... what exactly is $$h$$ ... clearly I need to understand the nature of h to understand $$f^*$$ ... ... 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 ... ... $$f \ : \ \text{Mod}_R \longrightarrow \text{Mod}_R \ \ \text{ where } \ \ f \ : \ M \longrightarrow N$$.Now we need $$f$$ to map to $$f^*$$ where $$f^* \ : \ \text{Ab} \longrightarrow \text{Ab} $$ ... ... So an $$f^*$$ defined by

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

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


So we have

$$f \ : \ M \longrightarrow N$$ and $$h \ : \ N \longrightarrow X$$

so then ...

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

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