Newbie into categorical proofs Let $$ F:Mod_A \to Mod_B $$ an aditive , exact  and covariant  functor  and  $$ M ∈ Mod_A $$ and $$M_1 , M_2 $$ submodules of  M . Show that $$ F( M_1\cap M_2)=F(M_1)\cap F(M_2 ) $$
$$ F( M_1+ M_2)=F(M_1)+ F(M_2 ) $$
As the title says so im  a newbie into categorical proofs so how im  supposed to proove this , probably not proving the double contention as set theory??
 A: You could ask yourself first: How this is meant at all?
What guarantees, that if $M_1\,\subseteq\,M$ is a submodule, then $F(M_1)$ is a submodule (or subset at all) of $F(M)$? 
 Nothing. But then how is the intersection or sum of $F(M_1)$ and $F(M_2)$  meant?
Well, we have the embedding $j_1:M_1\hookrightarrow M$ and $F$ maps it to $F(j_1):F(M_1)\to F(M)$. 
Use exactness to prove that $F(j_1)$ is also an embedding, i.e. injective, then $F(M_1)$ might be identified with the image of $F(j_1)$ which is already a flesh-and-blood submodule of $F(M)$.
Anyway, the pullback of the inclusions $j_1$ and $j_2$ will give their intersection, and again by exactness, $F$ preserves pullbacks. 
One can also define abstract subobjects as embeddings, partially ordered by $\underset{L\to M}j\ \ \le\, \underset{L'\to M}{j'}$ iff $\exists i:L\to L'\ $ s.t. $\ j=j'\circ i$.
Then the pullback as intersection can be rephrased as the infimum of the subobjects $j_1$ and $j_2$ w.r.t. this partial order, and then the sum is the supremum.
