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I struggle doing these type of questions, i have no idea how i should approach these type of question. Like for example, the question below:

Suppose $f: A\to B$ is an injective function. Prove that $f^{-1}(f(C))=C $ for all $C \subseteq A$.

I now know that I should define everything that is in the question and then im not sure what do next.

The definitions that i would use is the image of $f$ and inverse image of f(C) where $C\subseteq A$ and $f(C)\subseteq B$ and $f:A\to B$ is $f(C)$ $=$ ${\exists(c \in C) : f(c) = b}$.

The definition of injectivity is if $a_{1}$ != $a_{2}$ then $f(a_{1})$ != $f(a_{2})$.

Please tell me how you would approach these type of question. So I can do the same approach for other questions.

cosygod
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    Usually the standard way to show equality of sets is showing a two sided inclusion. Take an element in one of the sets and show it must belong to the other set, and vice versa. If you know the definitions, you should be able to do that. There is really nothing tricky about this. – Mark Dec 07 '23 at 11:00
  • @Mark yh it`s crazy, i dont know how to explain it, I just like unintentially jump to the conclusion and struggle to show the logic behind it. But i will try. – cosygod Dec 07 '23 at 11:06
  • cosy, Would you write out the definitions? – nickalh Dec 07 '23 at 11:23
  • @nickalh usually no, but other people say i have to so now i write out the definition. – cosygod Dec 07 '23 at 11:42
  • Again, please type out your definitions.

    A. Writing out the definitions is pretty standard for proofs at this level. Plus many proofs on this level are 75% givens, meaning stuff we can assume definitions. B. I'm working on an answer for you, but would like to copy and paste your definitions into my proofs. I feel like I'm working harder on your exercise than you are. C. Using your specific definitions will make it easier for you to understand as opposed to some similar definitions, which you may or may not be fully comfortable with.

     – nickalh Dec 07 '23 at 11:48
  • @nickalh ah i misunderstood your question. sure, here is the definition im going to use, i will type it properly up on the post because im not sure if it works here. f(C) is a set where there exists c in C such that f(c) = b} then for the set f(C) and a function f: A to B the inverse mapping of f(C) under f is $f^{-1}(f(C)) which is the set of all a in A such that f(a) in D. – cosygod Dec 07 '23 at 11:55
  • @nickalh I have presented the proper definitions that you can copy and paste, I believe i do not need to present the definition of injectivity, if its required for this proof, then I will create a definition for that too. – cosygod Dec 07 '23 at 12:05
  • @Mark Effective education: To any person who has done few or no proofs of these types, these exercises may not be simple at all. I think many trying to help forget the challenge it took to learn a particular type of proof or any math type of math exercise. Even if one type of exercise has always come easily to one student, basic sympathy or even empathy is helpful. – nickalh Dec 07 '23 at 13:21
  • Statements like "nothing tricky about this" and "you should be able to do this" are more likely to hinder learning than help learning by making the learner discouraged or even feel stupid. Effective teaching refrains from "this is simple" comments, especially when the learner has stated "I struggle doing these type of questions." – nickalh Dec 07 '23 at 13:21
  • I apologize @cosygod I had my answer mostly written before I saw your definitions. – nickalh Dec 07 '23 at 13:40
  • @nickalh ah no worries, your answer was understandable and useful, thank you for the help. – cosygod Dec 07 '23 at 15:02

1 Answers1

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Suppose $f: A\to B$ is an injective function. Prove that $f^{-1}(f(C))=C $ for all $C \subseteq A$.
Proof:
Part I will show $C \subseteq f^{-1}(f(C))$.
Let C be any subset of A.
For any $c \in C$ there exists, $b \in B$ such that $b = f(c).$
Because f is injective, $f^{-1}(b)$ has exactly one value, namely c.
This means $c \in f^{-1}(f(C))$
So $C \subseteq f^{-1}(f(C)) $

Proof Plans
A proof plan is an underused technique for teaching how to develop proofs. Proof plans are similar to rough drafts of an essay. They have enough details to see the overall structure and flow the proof will take. However, they do not need to get bogged down in technical details. Proof plans should not be included in the final proof, because they frequently have gaps. A proof plan for part II, could look like the following.

Proof Plan for Proving a Subset
A proof plan for most subset proofs, $D \subseteq E$, follows this pattern.
Let $d \in D$.
Prove $d \in E$.
Conclude $D \subseteq E$.

Proof Plan for Part II
Part II should show $f^{-1}(f(C)) \subseteq C$.
Assume $a \in f^{-1}(f(C))$.
$b \in f(C),$ such that $a \in f^{-1}(b)$.
Show a is distinct. Hint: f is injective. See note 2 below. $b \in f(C)$ and $b=f(a)$
$a \in C$
$f^{-1}(f(C)) \subseteq C$.

Depending on your professor, many details, particularly reasons and a conclusion still need to be fleshed out. The intention is simply to demonstrate how to approach these types of proof.

Note how both directions primarily use definitions such as definition of inverse, definition of a subset, definition of injective and definition of a function are all used or implied. Also, note how most of the lines rely on the immediate previous line in some way. For this level of proofs, it frequently works out this way, in part because there aren't a whole lot of theorems or definitions yet which will move the proof towards the ultimate goal.

Note 2: One subtle, possibly tricky point to be aware of. Injectivity is required for this theorem to be true.
Consider the noninjective function,
$f(1) =1, f(2) =1$, and the set $C = \lbrace 1 \rbrace $
$f^{-1}(1)=\lbrace 1, 2 \rbrace $
However, $\lbrace 1, 2 \rbrace \not\subset \lbrace 1 \rbrace$

In general, get to know the patterns for various types of proofs.
Existence,
Nonexistence- usually by contradiction or contrapositive.
Implicaton
Implication with negatives- frequently uses the contrapositive. etc.

nickalh
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