Given two arbitrary $R$ modules $M_1, M_2$, can we ask if $M_1 \subset M_2$? As title indicates, given ANY arbitrary $R-$modules can we compare the modules by inclusion? 
The question came up during a proof about Baer's Criterion in which in checks when a module $Q$ is injective. The statement follows

The module $Q$ is injective $\iff$ for all ideal $I\subset R$, any $R-mod$ homomorphism $g:I\to Q$ can be extended to $G:R \to Q$.

Now for the $\Longleftarrow$ direction, it invokes Zorn's Lemma in the following way. 
First given an exact sequence $0 \to L \stackrel{\phi}\to M$ and $f: L \to Q$, we need to extend to $F: M \to Q$.
It is invoked by declaring $S$ to be the set of lifts of $f$, that is $(f',L')$ such that


*

*$L'$ is a submodule such that $L \subset L' \subset M$

*$f'\vert_{L} = f$
Now what I don't understand is that if we can actually assume there is relation between $L$ and $M$? Do they actually mean to say $L \stackrel{\phi}\approx \phi(L) \subset L' \subset M$? 
I was under the impression that the definition of an $R-mod$ we can take $M$ to be any set.
You may find a reference of this proof in Dummit/Foote page 396.
 A: There are a few different answers to this, and the underlying issue is basically what it means for things to be "equal." In a strict sense, there are isomorphic modules which are not equal. For example, the $\Bbb Z$ module $\Bbb Z$ is isomorphic to $\Bbb Z[x]/(x)$, but these are not equal because elements of $\Bbb Z$ are numbers and elements of $\Bbb Z[x]/(x)$ are cosets.
Of course, some modules are truly contained in one another. The module $\Bbb Z$ is contained in $\Bbb Z[x]$ because any integer $n$ truly also lives in $\Bbb Z[x]$. In general, two arbitrary modules will not be comparable with respect to inclusion, i.e., we have both $A\not\subseteq B$ and $B\not\subseteq A$. So one can always ask whether $A\subseteq B$, but the answer will usually be no.
That said, people sometimes casually refer to one set "including" into another even when are is no actual inclusion. The point is that, morally, an can be replaced with any structure preserving injection. For example, we have an injection $i\colon \Bbb Z\to \Bbb Z[x]/(x)$ where $i(n)=n+(x)$. Some people will then say that we can now "identify" $\Bbb Z$ with its image under this map, or that "under this identification," $\Bbb Z$ is a submodule of $\Bbb Z[x]/(x)$.
In the example you reference, the map $\phi\colon L\to M$ is injective, which means that we can "identify" a point $\ell\in L$ with $\phi(\ell)$ in $M$. So by a small abuse of notation, we can just say $L\subseteq M$. If you want to be more careful, you could instead factor as an isomorphism from $L\to L'\subseteq M$, where $L'$ really is a submodule.
The moral is that we usually don't actually care that much about distinguishing between two isomorphic things, so we can afford to be a bit casual. More generally, a structure preserving injection from one thing to another is an "embedding," and the idea is that replacing 'strict inclusion' with 'embedding' turns out to be what we want.
