JEST 2020

Question

X AND Y is an arbitrary sets, F: $X\rightarrow Y$ show that a and b are equivalent 

1. F is one-one
2. For all set Z and function g1: $Z\rightarrow X$ and g2: $Z\rightarrow X$,  if $g1 \neq g2$ implies $f \bigcirc g1 \neq f \bigcirc g2$  Where $\bigcirc$ is a fucntion composition.

Comments

Kindly update the question....It was asked to show that (a) and (b) are equivalent.

Answer 1

 

A function f: A $\rightarrow$ B is said to be one-one if different elements of A have different images in B.
Thus, f is one-one 
$\Leftrightarrow$   a$\neq$b        $\Rightarrow$   f(a)$\neq$ f(b)   $\forall$ a,b $\in$ A
$\Leftrightarrow$   f(a)=f(b) $\Rightarrow$   a=b            $\forall$ a,b $\in$ A

 Let z$_{1}$, z$_{2}$ $\in$ Z such that x$_{1}$ = g$_{1}$(z$_{1}$) and x$_{2}$ = g$_{2}$(z$_{2}$) and x$_{1}$, x$_{2}$​​​​​​ $\in$ X.​

As it is already given in the question, statement b follows

 

if g1≠g2 implies f◯g1≠f◯g2

 we can rewrite it as 

$\Leftrightarrow$  g$_{1}$(z$_{1}$) $\neq$ g$_{2}$(z$_{2}$)  $\Rightarrow$  f(g$_{1}$(z$_{1}$)) $\neq$ f(g$_{2}$(z$_{2}$))

$\Leftrightarrow$  x$_{1}$ $\neq$ x$_{2}$ $\Rightarrow$   f(x$_{1}$) $\neq$ f(x$_{2}$)

Which is basically the definition of one-one function. So both statements a and b are equivalent.