Here is the question I want to solve:
Let $M_i (i \in I)$ be any family of $A$-modules, let $M$ be their direct sum. Prove that $M$ is flat $\iff$ each $M_i$ is flat.
Here is a solution to it found on the internet:
But I do not understand the following:
1-Why we need to identify $j \otimes id_M$ with a map $h$? also, what criteria should we use for choosing this map?
Here are the isomorphisms of (2.14 iii*):
And here is (2.22.i*):
The takeaway of Proposition 2.22 is that exactness can be checked coordinate wise, in the sense that if a short sequence $\mathsf S: 0\to M'\to M\to M''\to 0$ of morphisms is obtained by taking the direct sum of several short sequences of morphisms $0\to M_i'\to M_i\to M_i''\to 0$, then $\mathsf S$ is exact if and only if each summand sequence is itself exact.
Going to the second-to-last claim, this is saying that if one takes a sequence $\mathsf S$ and if $M$ is the sum of modules $M_i$, then $\mathsf S\otimes_ AM$ is the sum of the sequences $\mathsf{S}_i = \mathsf{S}\otimes_A M_i$. This means that $\mathsf S\otimes_ AM$ is exact if and only if each $\mathsf{S}_i = \mathsf{S}\otimes_A M_i$ is exact by the first paragraph.
Finally, $\mathsf{S}_i = \mathsf{S}\otimes_A M_i$ is exact for any $\mathsf{S}$ if and only if $M_i$ is flat (by definition, if you would like to) which means that $\mathsf{S}\otimes_A M$ is exact for any $\mathsf{S}$ if and only if each $M_i$ is flat. Thus $M$ is $A$-flat if and only if each $M_i$ is flat.
The map $h$ is simply the unique map that sends a finite sum $\sum_{i\in I}n_i\otimes m_i$ with $n_i\in N'$ and $m_i\in M_i$ to the finite sum $\sum_{i\in I} j(n_i)\otimes m_i$ where now $j(n_i)\in N$. There is a typo in the solutions and the $N'$ in the codomain should be an $N$. The whole motto of this result is that tensor products commute with direct sums: for any family of modules $\{M_i : i\in I\}$ and any module $N$, there is a natural isomorphism
$$ M\otimes N \longrightarrow \bigoplus_{i\in I} M_i\otimes N $$
where $M = \left(\bigoplus_{i\in I} M_i\right)$ that sends an element $m\otimes n$ with $m = \sum_{i\in I} m_i $ to $\sum_{i\in I} m_i\otimes n$. The inverse map sends an element $\sum_{i\in I} m_i\otimes n_i$ to the the sum $\sum_{i\in I} m_ie_i\otimes n_i$ where $m_ie_i\in M$ is the element that is zero everywhere except on the $i$th coordinate where it equals $m_i$.