A property of the tensor product of modules in Dummit and Foote's Abstract Algebra The following theorem is from the Abstract Algebra by Dummit and Foote (in the section 10.4 tensor products of modules):

Would anybody illustrate how Theorem 8 is used to get
$$
\textrm{ker }\iota\subset\textrm{ker }\varphi
$$
in the proof?  


 A: The second sentence of the proof of Corollary 9 is kind of poorly phrased.  Here's a clearer phrasing:

Suppose now that $L$ is an $S$-module, $K\subseteq N$ is a submodule, and $\psi:N/K\to L$ is an injective $R$-module homomorphism.  Write $\varphi:N\to L$ for the composition of $\psi$ with the quotient homomorphism $N\to N/K$, so $\ker(\varphi)=K$.

That is, the map $\varphi$ is exactly a map satisfying the hypotheses of Theorem 8, and $K$ is its kernel (where what we know about $K$ is that the quotient $N/K$ is a quotient that embeds into an $S$-module).  So let $\Phi:S\otimes_R N\to L$ be the map provided by Theorem 8 using the map $\varphi$.  Then for each $n\in N$, $\varphi(n)=\Phi(\iota(n))$.  In particular, if $n\in \ker(\iota)$, then $\varphi(n)=\Phi(0)=0$, so $n\in\ker(\varphi)$.
A: Put the maps into the diagram. Now take anything in $\ker \iota$, say $n$. By the diagram in Thm 8, it gets sent to $0$ in $S \otimes N$ which, since $\Phi$ is a homomorphism, then gets mapped to $0$ in $L$. Thus, by commutativity of the diagram, so then does $\varphi (n) = 0$ meaning that $n \in \ker \varphi$. 
