Exact Sequences - extending or lifting homomorphisms

[Math Amateur]

Dummit and Foote open their section (part of section 10.5) on projective modules as follows:View attachment 2463D&F then deal with the issue of obtaining a homomorphism from D to M given a homomorphism from D to L and then move to the more problematic issue of obtaining a homomorphism from D to M given a homomorphism from D to N. (Strangely they refer to N as "the quotient N?). The relevant text reads as follows:

View attachment 2464
D&F then give an example ... and my question pertains to this example ... the example reads as follows:
View attachment 2465In this example, D&F make the following statement:

"Any homomorphism \(\displaystyle F\) of \(\displaystyle D\) into \(\displaystyle M = \mathbb{Z} \)must map \(\displaystyle D\) to \(\displaystyle 0\) (since \(\displaystyle D\) has no elements of order \(\displaystyle 2\))"

Can someone please explain why this statement is true?

I am aware that isomorphisms map elements of a given order onto elements of the same order, but here we are only dealing with a homomorphism.

Also \(\displaystyle 0\) does not have order \(\displaystyle 2\) anyway!

Can someone please clarify these issues?

Peter

 

[Deveno]

It appears that when D&F say "order" they mean "additive order".

Now, while it is true that homomorphisms of additive groups do not PRESERVE order, the image's order always DIVIDES the order of any pre-image. Since 2 is prime, if an image of an element of order 2 from $\Bbb Z/2 \Bbb Z$ (that is to say 1, the only element which HAS order 2) does not have order 2, it must have order 1 (that is, is the additive identity).

So the only possible abelian group homomorphism (that is, $\Bbb Z$-module homomorphism) $\Bbb Z/2 \Bbb Z \to \Bbb Z$ is the 0-map (the integers under addition only contains ONE finite cyclic subgroup, the trivial one).

Try to keep in mind that modules are "mostly abelian groups" with a little ring action mixed in for spice. The addition is the DOMINANT operation (we'll pretend we don't know about tensor products for this discussion, where the scalar multiplication makes a BIG difference).