# Homework Help: Range of mappings proof

1. Apr 16, 2016

### Incand

1. The problem statement, all variables and given/known data
Let $f:S\to T$ be a given function. Show the following statements are equivalent:
a) $f$ is 1-1
b) $f(A\cap B) = f(A) \cap f(B),\; \forall A,B \in S$
c) $f^{-1}(f(A)) = A,\; \forall A \subseteq S.$

2. Relevant equations
Definition:
$f$ is 1-1 of $A$ into $B$ provided that $f(x_1) \ne f(x_2)$ whenever $x_1 \ne x_2, \; \; \; x_1,x_2 \in A$.

Definitions:
Let $f$ is a mapping $f:A \to B$:
If $E \subseteq A$ then $f(E)$ is the set of all elements $f(x)$ with $x \in E$.
If $E \subseteq B$ then $f^{-1}(E)$ denotes the set of all $x\in A$ such that $f(x) \in E$.

3. The attempt at a solution
I think I'm able to prove a) $\Longrightarrow$ b) and a) $\Longrightarrow$ c) but I can't complete the rest.
Lets first prove the general statement $A \subseteq f^{-1}(f(A))$ :
Take $\alpha \in A$ then $f(\alpha) \in f(A)$ and hence $\alpha \in f^{-1}(f(A))$.

We can also prove that $f(A \cap B) \subseteq f(A) \cap f(B)$:
Take $\alpha \in f(A \cap B)$ that means $\alpha = f(z)$ for some $z\in A \cap B$ and hence $\alpha \in f(A)\cap f(B)$.

It's left to prove the equivalence between
a) $f$ is 1-1
b) $f(A) \cap f(B) \subseteq f(A\cap B),\; \forall A,B \in S$
c) $f^{-1}(f(A)) \subseteq A,\; \forall A \subseteq S.$
a) $\Longrightarrow$ b)
Take $\alpha \in f(A) \cap f(B)$ then $\alpha = f(z_1), \; z_1 \in A$ and $\alpha = f(z_2), \; z_2 \in B$. But since $f$ is 1-1 $z_1 = z_2$ hence $\alpha \in f(A \cap B)$ and $f(A) \cap f(B) \subseteq f(A\cap B)$.

a) $\Longrightarrow$ c)
Take $\alpha \in f^{-1}(f(A))$ that is $z = f(\alpha)$ for some $z\in B$. That is
$f(\alpha) \in f(A)$ hence $f(\beta) = z$ for some $\beta \in A$ but since $f$ is 1-1 this means $\alpha = \beta$ and $\beta \in A$ so $f^{-1}(f(A)) \subseteq A$.

To complete the proof I need to either show that c) $\Longrightarrow$ a) and b) $\Longrightarrow$ c) OR show that c) $\Longrightarrow$ a) and b) $\Longrightarrow$ a).

c) $\Longrightarrow$ a)
It's equivalent to show the contrapositive that $f(x_1) = f(x_2) \Longrightarrow x_1 = x_2$. Take $x_1, x_ 2 \in A$ so that $f(x_1)= f(x_2)$ then by c) $x_1,x_2 \in f^{-1}(f(A))$. This means that $z_1 = f(x_1)$ and $z_2 = f(z_2)$ for $z_1,z_2 \in B$ but from the premise $z_1 = z_2$.

I don't seem to get anywhere with the last part nor any luck with any of the other equivalences. Any hints on how to go about it? I'm also wondering If what I've done so far is correct?

2. Apr 16, 2016

### Samy_A

a) ⇒ b) and a) ⇒ c) are correct.
I don't exactly understand what you did in c) ⇒ a)

Hint for b) ⇒ c)
Take $A \subset S$, and set $B=f^{-1}(f(A)) \setminus A$. Use b) to prove that $B= \varnothing$.

Hint for c) ⇒ a)
Take $x\in S$ and apply c) to $A=\{x\}$.

3. Apr 16, 2016

### Incand

Cheers! The hints really helped! Think I got them now.

b) $\Longrightarrow$ c)
Let $A \subseteq S$ and set $B = f^{-1}(f(A))\backslash A$ then $A \cap B = \varnothing$. Using b)
$f(\varnothing ) = f(A \cap B) = f(A)\cap f(B), \; \; \forall A \subseteq S$. Hence
$f(\varnothing ) =f(B)$ but $f(\varnothing) = \varnothing$ and $f(B) = \varnothing$ only when $B = \varnothing$ so $B=\varnothing$.
This gives us that $f^{-1}(f(A)) = A, \; \; \forall A\subseteq S$.

c) $\Longrightarrow$ a)
Take $x\in S$ and take $A = \{x\}$ then by c) $f^{-1}(f(A)) = A= \{x\}$. Since the inverse image of $f(A)$ has only one element there is only one $x$ satisfying $z=f(x)$ for each $x\in S$. That means if $x_1 \ne x_2$ $f(x_1) \ne f(x_2)$ and hence $f$ is 1-1.