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-tak

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Question Number 156128 by aaaspots last updated on 08/Oct/21

$$f:X\to Y$$$$f\left(E\mathrm{\setminus }F\right)=f\left(E\right)\mathrm{\setminus }f\left(F\right)\Rightarrow fis1to1$$$$$$$$Ithinkitisnottrue$$$$sinceletx1x2x3\in E,x3x4\in F,$$$$andf\left(x1\right)=f\left(x2\right)itwillbenottrue.$$$$$$$$butmyfriendsayby\sim q\Rightarrow \sim p$$$$fisnot1to1$$$$\Rightarrow f\left(E\mathrm{\setminus }F\right)\ne f\left(E\right)\mathrm{\setminus }f\left(F\right),andtake$$$$x1x2\in X,f\left(x1\right)=f\left(x2\right)=y0$$$$E=\{x1,x2\}F=\left\{x2\right\}$$$$andcanproofitistrue$$$$butIdonotknowwhichistrue$$$$howtoproofit?$$

Answered by TheSupreme last updated on 08/Oct/21

$$$$$$D=\{y\mid y=f\left(x\right),x\in E\}$$$$C=\{y\mid y=f\left(x\right),x\in F\}$$$$if\mathrm{\forall }x\in Efis{}^{\prime }1to{1}^{\prime }\left(injective\right)and\mathrm{\forall }x\in inFf\left(x\right)isinjectiveand$$$$\mathrm{\forall }x\in E,y\in Ff\left(x\right)\ne f\left(y\right)so$$$$D\mathrm{\setminus }C=\{y\mid y=f\left(x\right),x\in E\mathrm{\setminus }F\}$$$$$$$$dimassuming\nexists x\in F,y\in E:f\left(x\right)=f\left(y\right)$$

Commented by aaaspots last updated on 08/Oct/21

$$Istilldonotunderstand.$$$$andcanweassumefis1to1$$$$whenweproofitatfirst?$$$$$$

Commented by aaaspots last updated on 08/Oct/21

$$welliwritef\left(E\mathrm{\setminus }F\right)=f\left(E\right)\mathrm{\setminus }f\left(F\right)$$$$onlyiffis1to1$$

Commented by TheSupreme last updated on 08/Oct/21

$$f\left(x\right)is1to1onlyiftheproposition\left(1\right)istrueforallE,F\in D\left[f\left(x\right)\right]$$$$Disdominium,prop1isyourquestion$$

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