f:X→Y f(E\F)=f(E)\f(F)⇒f is 1 to 1 I think it is not true since let x1 x2 x3∈E ,x3 x4∈F, and f(x1)=f( x2) it will be not true. but my friend say by ∼q⇒∼p f is not 1 to 1 ⇒f(E\F)≠f(E)\f(F),and take x1 x2∈X, f(x1)=f(x2)=y0 E={x1 ,x2} F={x2} and can proof it is true but I do not know which is true how to proof it?


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Question Number 156128 by aaaspots last updated on 08/Oct/21
$f:X→Y f(E\F)=f(E)\f(F)⇒f is 1 to 1 I think it is not true since let x1 x2 x3∈E ,x3 x4∈F, and f(x1)=f( x2) it will be not true. but my friend say by ∼q⇒∼p f is not 1 to 1 ⇒f(E\F)≠f(E)\f(F),and take x1 x2∈X, f(x1)=f(x2)=y0 E={x1 ,x2} F={x2} and can proof it is true but I do not know which is true how to proof it?$
$f : X \to Y$ $f (E ∖ F) = f (E) ∖ f (F) \Rightarrow f i s 1 t o 1$ $I t h i n k i t i s n o t t r u e$ $s i n c e l e t x 1 x 2 x 3 \in E, x 3 x 4 \in F,$ $a n d f (x 1) = f (x 2) i t w i l l b e n o t t r u e .$ $b u t m y f r i e n d s a y b y \sim q \Rightarrow\sim p$ $f i s n o t 1 t o 1$ $\Rightarrow f (E ∖ F) \neq f (E) ∖ f (F), a n d t a k e$ $x 1 x 2 \in X, f (x 1) = f (x 2) = y 0$ $E = {x 1, x 2} F = {x 2}$ $a n d c a n p r o o f i t i s t r u e$ $b u t I d o n o t k n o w w h i c h i s t r u e$ $h o w t o p r o o f i t ?$
	Answered by TheSupreme last updated on 08/Oct/21
	$D={y∣ y=f(x) , x∈E} C={y∣ y=f(x), x∈F} if ∀x∈E f is ′1 to 1′ (injective) and ∀x∈inF f(x) is injective and ∀x∈E,y∈F f(x)≠f(y) so D\C={y∣ y=f(x), x∈E\F} dim assuming ∄x∈F,y∈E: f(x)=f(y)$
	$D = {y ∣ y = f (x), x \in E}$ $C = {y ∣ y = f (x), x \in F}$ $i f \forall x \in E f i s^{'} 1 t o 1^{'} (i n j e c t i v e) a n d \forall x \in i n F f (x) i s i n j e c t i v e a n d$ $\forall x \in E, y \in F f (x) \neq f (y) s o$ $D ∖ C = {y ∣ y = f (x), x \in E ∖ F}$ $d i m a s s u m i n g ∄ x \in F, y \in E : f (x) = f (y)$
	Commented by aaaspots last updated on 08/Oct/21

	$I s t i l l d o n o t u n d e r s t a n d .$ $a n d c a n w e a s s u m e f i s 1 t o 1$ $w h e n w e p r o o f i t a t f i r s t ?$
	Commented by aaaspots last updated on 08/Oct/21
	$well i write f(E\F)=f(E)\f(F) only if f is 1 to 1$
	$w e l l i w r i t e f (E ∖ F) = f (E) ∖ f (F)$ $o n l y i f f i s 1 t o 1$
	Commented by TheSupreme last updated on 08/Oct/21

	$f (x) i s 1 t o 1 o n l y i f t h e p r o p o s i t i o n (1) i s t r u e f o r a l l E, F \in D [f (x)]$ $D i s d o m i n i u m, p r o p 1 i s y o u r q u e s t i o n$
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