Image of f under $A \subseteq X$ confusion and exercise So, I am reading the book Algebra: Abstract and Concrete for an upcoming Algebra 1 course in a couple weeks. I am starting with some preliminary material and have found myself a bit confused when reading about the concept of Image of A under f. The book defines the concept thusly:

If $f: X \rightarrow Y$ is a function and $A$ is a subset of $X$ we write $f(A)$ for {$f(a): a \in A$}. Thus $y \in f(A)$ iff there exists an $a \in A$ such that $f(a) = y$

So, upon first look my interpretation of what this definition is saying is that:

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*The image of A under f, f(A), is the set of values that includes the "outputs" of the set $A \subseteq X$ under the function f and an element of the set Y (y) is an "output" of this function if we have an element of A (a) that maps to the element $y \in Y$.

*This is simultaneously obvious and confusing to me at once. It seems obvious because it's basically saying that we have a subset A of a set X that maps to another set Y through the function f and if an element of Y is mapped to by the function f from an element of this subset A then that element y appears in the set known as the image of f under A. This just seems a little redundant and I'm almost wondering why it needs its own definition.

*A little clarification here or maybe a concrete example would be very helpful!


Next I have an exercise that pertains to this topic that I am a little stuck on. The exercise states:

Let $f: X \rightarrow Y$ be a function and let $E$ and $F$ be subsets of $X$. Show that $f(E) \cup f(F) = f(E \cup F)$

So for this my understanding of the objective is that we want to show that $f(E) \cup f(F) \subseteq f(E \cup F)$ and also that $f(E \cup F) \subseteq f(E) \cup f(F)$ in order to show equivalence between the 2 sets. I was going to start by saying that if $y \in f(E) $ or $y \in f(F)$ and then use for facts to try to imply that the element $y$ would be an element of $f(E \cup F)$ but I'm not sure how to do it exactly.

*

*Maybe am I supposed to start with elements, say $e$ and $f$ and then suing the def. of the image of f under E or F say that if $e \in E$ or $f \in F$ then that implies $f(e) \in f(E)$ or $f(f) \in f(F)$ and go from there? I'm quite confused, any help would be greatly appreciated!

 A: I'm not sure what you mean by your "redundancy" comment. $f(A)$ basically means the set of outputs where the inputs are from amongst $A$.
Regarding your exercise: As you mentioned, we should show $f(E) \cup f(F) \subseteq f(E \cup F)$ and $f(E \cup F) \subseteq f(E) \cup f(F)$.
So for the first one: suppose $y \in f(E) \cup f(F)$. By the definition of union this means that $y \in f(E)$ or $y \in f(F)$. First suppose $y \in f(E)$. That means there is an $x \in E$ such that $f(x) = y$. But if $x \in E$ then $x \in E \cup F$ which means that, since $f(x) = y$ and $x \in E \cup F$, $f(x) = y \in f(E \cup F)$. Now suppose $y \in f(F)$  instead. An identical argument would again lead you to the conclusion that $y \in f(E \cup F)$. Thus, in either case ($y \in f(E)$ or $y \in f(F)$) you come to the same conclusion that $y \in f(E \cup F)$, and so $f(E) \cup f(F) \subseteq f(E \cup F)$.
Now we show $f(E \cup F) \subseteq f(E) \cup f(F):$ But it's essentially the same argument. Suppose $y \in f(E \cup F)$; then there is some $x \in E \cup F$ such that $f(x) = y$. Now we have two options: either $x \in E$ or $x \in F$. First suppose $x \in E$. Then since $x \in E$ and $f(x) = y$ this means that $y \in f(E)$. An identical argument for the case that $x \in F$ would show that $y \in f(F)$. So what we have now is that: if $y \in f(E \cup F)$, then $y \in f(E)$ or $y \in f(F) \implies f(E \cup F) \subseteq f(E) \cup f(F)$.
A: Your view of the image of a set under a function is exactly right. Some intuition may be that it is essentially the range of the function. If we write a function $f:A\rightarrow B,$ then it is assumed that the domain of the function is $A$, but the range of the function need not be the entire set $B$, so we have a notion of the image, which satisfies $f(A) \subseteq B$. A function is surjective, for example, if $f(A)=B.$ A lot of math is done using sets, so the notion of an image lets us talk about all the points mapped to under the function $f$ as a set. Its utility will become more clear as you continue through your courses.
Your approach to the proof is correct. Take some $y \in f(E) \cup f(F)$. This means $y \in f(E)$ or $y\in f(F)$. Now by definition of the image, what does this mean?
