A Universe of Sorts

An injective module is a generalization of the properties of $\mathbb Q$ as an abelian group ( $\mathbb Z$ module.)
In particular, given any injective group homomorphism $f: X \to Y$ and a morphism $q_X: X \to \mathbb Q$ , then we induce a group homomorphism $q_Y: Y \to \mathbb Q$ , where $X, Y$ are abelian groups.
We can think of this injection $f: X \to Y$ as identifying a submodule (subgroup) $X$ of $Y$ .
Suppose we wish to define the value of $q_Y$ at some $y \in Y$ . If $y$ is in the subgroup $X$ then define $q_y(y) \equiv q_x(y)$ .
For anything outside the subgroup $X$ , we define the value of $q_y$ to be $0$ .
Non-example of injective module: See that this does not work if we replace $\mathbb Q$ with $\mathbb Z$ .
Consider the injective map $Z \to Z$ given by $i(x) \equiv 3x$ Consider the quotient map $f: Z \to Z/3Z$ . We cannot factor the map $f$ through $i$ as $f = ci$ [ $c$ for contradiction ]. since any map $c: Z \to Z/3Z$ is determined by where $c$ sends the identity. But in this case, the value of $c(i(x)) = c(3x) = 3xc(1)) = 0$ . Thus, $\mathbb Z$ is not an injective abelian group, since we were unable to factor the homomorphism $Z \to Z/3Z$ along the injective $3 \times: Z \to Z$ .
Where does non-example break on Q? Let's have the same situation, where we have an injection $i: Z \to Q$ given by $i(z) = 3z$ . We also have the quotient map $f: Z \to Z/3Z$ . We want to factor $f = qi$ where $q: Q \to Z/3Z$ . This is given by $q(x) =$

§ Injective Module