Proof for the universal property of the tensor product of modules The following is the theorem regarding the universal property of tensor products of modules in Dummit and Foote:

Let $R$ be a subring of $S$, let $N$ be a left $R$-module and $\iota:N\to S\otimes_RN$ be the $R$-module homomorphism defined by $\iota(n)=1\otimes n$. Suppose that $L$ is any left $S$-module (hence also an $R$-module) and that $\varphi:N\to L$ is an $R$-module homomorphism from $N$ to $L$. Then there is a unique $S$-module homomorphism $\Phi:S\otimes_RN\to L$ such that $\varphi$ factors through $\Phi$, i.e., $\varphi=\Phi\circ\iota$ and the diagram
  
commutes. Coversely, if $\Phi: S\otimes_RN\to L$ is an $S$-module homomorphism then $\varphi=\Phi\circ\iota$ is an $R$-module homomorphism from $N$ to $L$. 

The proof goes as follows


*

*Define $\psi:S\times N\to L$ with $\psi(s,n)=s\varphi(n)$.

*By the universal property of free $\mathbf{Z}$-module on the set $S\times N$, denoted as $F(S\times N)$, there exists a $\mathbf{Z}$-module homomorphism $\Psi:F(S\times N)\to L$ such that $\Psi(s,n)=\psi(s,n)=s\phi(n)$.

*Since $\varphi$ is an $R$-module homomorphism, the generators of the subgroup $H$ in the following equations all map to zero in $L$, where the generators are given by elements of the form
$$
(x+y,n)-(x,n)-(y,n),\quad (x,m+n)-(x,m)-(x,n),\quad (sr,n)-(s,rn),\\
x,y\in S,\,m,n\in N,\, r\in R.
$$

*$\color{blue}{\textrm{Hence}}$, $\Psi$ factors through $H$, i.e., there is a well defined $\mathbf{Z}$-module homomorphism $\Phi$ from $F/H=S\otimes_RN$ to $L$ satisfying $\Phi(s\otimes n)=s\varphi(n)$.

*The rest of the proof can be read below. 


Here is my question:

Could anyone elaborate how the "Hence" step is implied by the third bullet point, from which I can only see that $H\subset \ker\Psi$?


Here is the original proof in the book:

 A: $H \subseteq \textrm{ker } \Psi$ is all you need for $\Psi$ to factor through the quotient $F/H$.
In general, let $\delta: A \rightarrow B$ be a homomorphism of abelian groups ($\mathbf Z$-modules, same thing), and let $A_0$ be a subgroup of $A$.  To say that $\delta$ factors through $A_0$ (or $A/A_0$) is to say that there exists an abelian group homomorphism $\bar{\delta}: A/A_0 \rightarrow B$ such that $\bar{\delta}(a + A_0) = \delta(a)$ for all $a \in A$.
If such a homomorphism exists, it is clearly unique.  The only issue barring the existence of such a homomorphism is the possibility that it is not well defined.  And you can check that it is well defined if and only if $A_0$ is contained in the kernel of $\delta$.
A: On page 100 of Dummit and Foote, there is a remark about the "universal property" of quotient groups, which explains the "Hence" step (in particular it explains what "factors through" means). 
One can also read this note: 
https://ocw.mit.edu/courses/mathematics/18-703-modern-algebra-spring-2013/lecture-notes/MIT18_703S13_pra_l_9.pdf
