What's wrong with my counter example on $f^{-1}(f(A))=A \iff f$ is injective?

Question

$f:X \to Y , A\subset X, B\subset Y$

$f^{-1}(f(A))=A \iff f$ is injective

My counter example is setting function $f$ and $A$ such that $A=\{1, 2\}, f(A)=\{a\}, f^{-1}(f(\{a\}))=\{1, 2\}$ which is setting preimage of f(A) to A

for me it seems such definition of subset and function holds

$f^{-1}(f(A))=A $

because by definition of preimage $\{1, 2\}=\{x\in X \vert f(x)=A \}$

and also non injectivity.

but my text book and another lecture insist above property is true.

Can someone find error in my counter expamle please?

Perhaps this will be useful to you: https://math.stackexchange.com/questions/261157/show-s-f-1fs-for-all-subsets-s-iff-f-is-injective — scoopfaze, Jan 25 '20 at 07:30

score 2 · Accepted Answer · answered Jan 25 '20 at 07:53

2

The statement is false. What is true is $f$ is injective iff $f^{-1}(f(A))=A$ for every set $A$ in the domain of $f$.

answered Jan 25 '20 at 07:53

Kavi Rama Murthy

311,013

OHHHHH MAA GAAA... I'm baka desu... small brain.. Thank you so much – WienAudience Jan 25 '20 at 07:55

score 0 · Answer 2 · answered Jan 25 '20 at 08:34

$$f^{-1}(f(A))=A \implies f(x)$$ is a bijection (one-one(injective) and onto): $f: A \rightarrow B$, where $A$ and $B$ are subsets of $X$ and $Y$. Also the bijectivity of $f(x): A \rightarrow B$ is both necessary and sufficient for $f^{-1}(x)$ to exist.

What's wrong with my counter example on $f^{-1}(f(A))=A \iff f$ is injective?

2 Answers2