Question about proof on left module structure of $M\otimes N$ Here is the screenshot of the proof on $M\otimes_S N$ being a left $R$-module when $M$ is an $(R,S)$-bimodule and $N$ a left $S$-module, taken from Module Theory by Blykh.

What I am wondering is why, instead of verifying that $M\otimes_S N$ is a left $R$-module under such action directly, the author defined a balanced map $\vartheta_r$ compatible with the action and induced a $\Bbb Z$-endomorphism $f_r$ on $M\otimes_S N$ from it.
Here are some of my explanations:

When we are settled down to verify the left $R$-module structure on $M\otimes_S N$, the axioms
\begin{align*}
(r+s)\cdot p&=r\cdot p+s\cdot p \\
(rs)\cdot p&=r\cdot(s\cdot p) \\
1_R\cdot p&=p
\end{align*}
where $p\in M\otimes_S N$ are rather trivial. Nonetheless, the last axiom for $p_1,p_2\in M\otimes_S N$ $$r\cdot(p_1+p_2)=r\cdot p_1+r\cdot p_2$$ is nontrivial. The definition of the action relies on a specific representation of each $p\in M\otimes_S N$. It would be hard for us to obtain the desired formula once the representation of $p_1+p_2$ is given differently from the representations of $p_1$ and $p_2$. However, for the sake of endomorphism $f_r$ on $M\otimes_S N$, everything is way simpler, as such axiom follows merely by the linearity of $f_r$.

Not sure if my intuitions are correct. Hope to hear from you guys.
Update: I also noticed that the function $f_r$ ensured the action to be well-defined. It is always a big problem when we define functions on elements that rely on specific structures.
 A: The reason for defining a map likes this comes even before checking the axioms of a module: You need to have a well-defined skalar multiplication $(r,p)\mapsto rp$ for a ring element $r$ and every element $p$ of the module. Since elements may have more than one representation in tensor products, it is not entirely obvious that the multiplication $(r,\sum_im_i\otimes_Sn_i)\mapsto\sum_i(rm_i)\otimes_Sn_i$ is well-defined.
To prove that a map from a tensor product is well-defined, you almost always show that the corresponding map from the product of the modules is bilinear, and thus induces a homomorphism from the tensor product by the universal property of the latter.
A: This is a really common type of proof when talking about the tensor product. The point is that you can get a lot done by utilizing the universal property of the tensor product, i.e. that a balanced map from the cartersian product induces a morphism from the tensor product.
The axioms of a module that you a verifying are the most standard ones, i.e. given in terms of an action of a ring on an abelian group. An equivalent set of axioms can be formulated in terms of a morphism from the ring into the endomorphism group of the underlying abelian group. That is what the author did here.
