I know this part of Atiyah and Macdonald has a typo, but that is not what this question is about.
Let $R$ be a commutative ring and $S$ and $T$ be $R$ algebras. I am trying to show that
$S\otimes_R T$ has the structure of an $R$ algebra. To do this I first want to show that
$S\otimes_R T$ is a ring, so I need a multiplication map. I start with defining the map:
$$
\begin{align*}
\phi:(S\times T)\times (S\times T)&\longrightarrow (S\otimes _R T)\\
((s_1,t_1),(s_2,t_2))&\longmapsto (s_1s_2)\otimes (t_1t_2)
\end{align*}
$$
which is clearly $R$ linear in each entry, hence by the universal property of tensor products (or I guess more precisely a consequence of this property) factors through a map:
$$
\begin{align*}
\psi:(S\otimes_R T)\otimes_R (S\otimes_R T)&\longrightarrow (S\otimes _R T)
\end{align*}
$$
which satisfies:
$$
\begin{align*}
\phi((s_1,t_1),(s_2,t_2))=\psi\circ \otimes^4((s_1,t_1),(s_2,t_2))
\end{align*}
$$
where $\otimes^4:(S\times T)\times (S\times T)\rightarrow
(S\otimes_R T)\otimes_R (S\otimes_R T)$ is the map such that:
$$
\begin{align*}
\otimes^4((s_1,t_1),(s_2,t_2))=s_1\otimes t_1\otimes s_2\otimes t_2
\end{align*}
$$
The above all makes sense to me, but then Atiyah and Macdonald state that by proposition $(2.11)$ $\psi$ corresponds to a bilinear map:
$$\mu:(S\otimes_R T)\times(S\otimes_R T)\longrightarrow (S\otimes_R T)$$
I do not understand this step at all. Namely, proposition $(2.11)$ is:
Let $0\rightarrow M_0\rightarrow M_1\rightarrow\cdots\rightarrow M_n\rightarrow 0$ be an exact sequence of $R$ modules in which all the modules $M_i$ and the kernels of all the homomorphisms are belong to a class of modules $C$. Then for any additive function $\lambda$ on $C$ we have:
$$ \sum_{i=0}^n(-1)^i\lambda(M_i)=0$$
I believe I understand the proof of this statement, but I just do not see how it applies in this case at all. Is there a different way to see that there is a bilinear map $\mu$ which corresponds to the map $\psi$? Any help would be appreciated.
Edit:
So Atiyah and Macdonald don't actually define a class of modules from what I can see, but the way it is used seems to imply it is simply a collection of $R$ modules.
An additive function on $C$ is then a function is then a function on $C$ with values in an abelian group such that for every short exact sequence:
$$0\rightarrow M_0\rightarrow M_1\rightarrow M_2\rightarrow 0$$
where each $M_i\in C$ we have:
$$\lambda(M_0)-\lambda(M_1)+\lambda(M_2)=0$$
Best Answer
That is an incorrect internal reference. It should say (2.12) instead of (2.11).
Anyway, setting $M = S \otimes_R T$, you are asking how a linear map $\psi \colon M \otimes_R M \to M$ corresponds to a bilinear map $\mu \colon M \times M \to M$. Do you see that this is simply the universal mapping property of tensor products? That is (2.12).
For another approach to showing the tensor product of two $R$-algebras has a unique $R$-algebra structure such that elementary tensors multiply in the natural way, see Theorem 7.1 here.