I just started to learn some multilinear algebra, that I lacked in my undergraduate program. I decided to do it with the book of Douglas Northcott.
He defines what is going to be the solution for the universal problem for multilinear mappings.
Given $R$ a commutative non-trivial ring which possesses an identity element, let $p\in \mathbb{N}\setminus\{0\}$ and $M_1,\dotsc,M_p$ be $R$-modules, he denotes $U(M_1,\dotsc,M_p)$ as the free ring generated by the cartesian product $M_1\times\dotsb\times M_p$.
The elements of $U(M_1,\dotsc,M_p)$ that have one of the following forms $$(m_1,\dotsc,m_i+m_i',\dotsc,m_p)-(m_1,\dotsc,m_i,\dotsc,m_p)-(m_1,\dotsc,m_i',\cdots,m_p)$$ or $$(m_1,\dotsc,rm_i,\dotsc,m_p)-r(m_1,\dotsc,m_i,\dotsc,m_p)$$ generate a submodule that he denotes as $V(M_1,\dotsc,M_p)$. Then he does $M:=U(M_1,\dotsc,M_p)/V(M_1,\dotsc,M_p)$ and defines the mapping $\phi:M_1\times\dotsb\times M_p\rightarrow M$ by letting $\phi(m_1,\dotsc,m_p)$ is the natural image of $(m_1,\dotsc,m_p)$ considered as an element of $U(M_1,\dotsc,M_p)$ in M. Since the elements that generate $V(M_1,\dotsc,M_p)$ become zero in M, $\phi$ is multilinear.
Then he goes onto proving that the universal problem for multilinear mappings has unique solution up to isomorphism. But then as a corollary he says that each element of $M$ can be expressed as a finite sum of elements of the form $\phi(m_1,\dotsc,m_p)$.
This is where the confusion to me lays, in his proof,
Suppose that $M'$ is the $R$-submodule of $M$ generated by elements of the form $\phi(m_1,\dotsc,m_p)$. Also let $h_1:M\rightarrow M/M'$ be the natural homomorphism and $h_2:M\rightarrow M/M'$ be the null homomorphism. Then $h_1\circ\phi=h_2\circ\phi$ and therefore $h_1=h_2$. However, this implies that $M=M'$.
The proof continues but here I have the problem that I don't see why is the case that $h_1\circ\phi=h_2\circ\phi$, but when I somehow see it is when I think that they have the same image for every element, but I think I am wrong in this, but if I am not then almost everything seems unnecessary as by construction then you have that $M=M'$, which is part of what he is trying to prove. I get the feeling that the kind of argument that he does is very common in homological algebra (maybe not, right now I am complete ignorant of that field).
I am sorry if this is very low level for the site, I feel kind of embarassed as I have been stucked in this for two days.