The definition is given here: Show the direct sum is a proper subset of the direct product.
1) I did not understand why For a finite number of factors, the direct sum and direct product of abelian groups (and more generally, of R-modules) are equal? could anyone explain this for me please?
2) But I do not understand why the domain of the function $f$ is $I$ and why its codomain is $\cup A_{i}$ in both definitions? could anyone explain this for me please?
Thanks!!
On
Your question is not about a theorem such as "why something happens ?", instead it is "Why it is defined that way?". With this interpretation (I can't think of any other) the simple answer is: They give rise to two different things: this means if we have only one variety then we will miss out significant something else.
Assume all the $$A_i$ are the same, viz., the real numbers. Thenn direct sum will represent a a sequence of numbers which has a tail of zeros and can be interpreted as the set of all polynomials with real coefficients.
The direct product will represent every sequence of real numbers which can be interpreted as (coefficients of) a power series.
In the former everything is a globally defined nice function continuous and differentiable everywhere.
In the latter, the example $(1,1,1,1,\ldots)$ will represent the power series $\frac1{1-x}$ which is a function with a discontinuity at origin.
SO direct product brings into its fold functions which are just local (defined just on an interval, for example).
In both the direct product and the direct sum, think of $f$ as a "sequence". The input values $i \in I$ are the "indices" of the sequence $f$. The $i$th entry of this sequence is $f(i)$ and it is required to be in the group $A_i$. Thus, $f(1) \in A_1$, $f(2) \in A_2$, and so on.
For example consider $I = \{1,2\}$. The two groups $A_1$ and $A_2$ are allowed to be entirely different groups: the first coordinate is an element of the group $A_1$; and the second coordinate is an element of the group $A_2$. An element of $A_1 \times A_2$ or of $A_1 \oplus A_2$ can be thought of as a sequence of length $2$, also known as an ordered pair, whose first coordinate is in $A_1$ and whose second coordinate is in $A_2$. Formally, we might prefer to write the element $f : \{1,2\} \to A_1 \cup A_2$ as an ordered pair $(f(1),f(2))$. Furthermore, we might prefer to write our ordered pairs in the form $(x_1,x_2)$. But this is a mere notational difference: write $f$ instead of $x$; and write the index between parentheses instead of as a subscript.
The difference between the direct product and the direct sum is that, for various reasons, one might want to put further restrictions on the coordinates of $f$. In the direct product $\prod_{i \in I} A_i$, there are no further restrictions on the coordinates of $f$.
But in the direct sum $\oplus_{i \in I} A_i$, there is a further restriction on $f$: all but finitely many coordinates must be the zero element. In other words $f$ must satisfy the following property: there exists a finite subset $J \subset I$ such that if $i \in I-J$ then $f(i)$ is the zero element of $A_i$.
If $I$ were a finite set then the direct sum definition and the direct product definition would result in the exact same set of functions.
However, in the case of an infinite index set such as $I = \mathbb{Z}_+$, we can easily write down an element $f$ which is in the direct product but not in the direct sum: simply assign $f(1)$ to be a nonzero element of $A_1$, and assign $f(2)$ to be a nonzero element of $A_2$, and generally for any index $i \in I$ assign $f(i)$ to be a nonzero element of $A_i$.