Easy ten points. I can't seem to get what seems to be a very obvious result? Given module homomorphism $\alpha:M\to N$ and $P\le M$ (submodule). Can we say $\alpha(P)\le \alpha(M)$? In the solution the lecturer points out that the image of $\alpha$ is the same as the image of $\alpha\circ i$ where $i$ is the inclusion map $i:P\to M$. How does this help us see $\alpha(P)\le \alpha(M)$? 
If the inclusion map is not relevant to showing $\alpha(P)\le \alpha(M)$, perhaps it's needed in some way to show $\alpha(P)\leq N $? (please see comment below) 

First, you need to show that the image of a module under a homomorphism is also a module.  Once you have shown that then $\alpha(P)$ is certainly a subset of $\alpha(M)$.  Also, $P$ and $M$ are both modules.  Also, $\alpha(P)$ and $\alpha(M)$ are both modules by the first step.
