Forgive me for this naive question. 


We consider the following lemma and its proof in Lang's algebra, Third Ed., published 1999, Chap. 20, section 4, [page 784](https://books.google.com/books?id=Fge-BwqhqIYC&pg=PA784).



> 
> Every module is a submodule of an injective module.
> 
> 
> 


For a module $M$, Lang defines its dual to be $M$^ = $Hom(M, \mathbb{Q}/\mathbb{Z})$.


This "dual", if anything, is an algebraic version of the Pontrjagin dual of a topological group $G$, which is $Hom(G, \bf{S}^1)$, where we consider the continuous homomorphisms. $\mathbb{Q}/\mathbb{Z}$ is the torsion part of $\bf{S}^1$, and it seems this restriction is done for preventing too big a dual.


The proof goes on and shows that any module is a submodule of an injective module.


However, is there any conceptual explanation of why this works, and why this mysterious appearance of a dual which looks so much like the Pontrjagin dual?



Injective modules and Pontrjagin duals
Why this works? Because $\mathbb{Q}/\mathbb{Z}$ is an [injective cogenerator](https://en.wikipedia.org/wiki/Injective_cogenerator) in the category of abelian groups.


The mysterious appearance can be explained if you look at it from a more general perspective. This would be Morita duality, which was motivated by classical Pontryagin duality (locally compact Hausdorff topological groups), and it replaces the role of the circle group with an injective cogenerator.


In the category Set of sets and mappings, for example, the two point set $\{0,1\}$ is an injective cogenerator. The category Gr of groups and group homomorphisms on the other hand doesn't have any cogenerator.


