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- Homework Statement
- to the set, is the function injective? Let ##f## be a function. Denote the domain ##A## and the target space ##B##. Let ##X\subset A##. If ##f^{-1}(f(X))=X##, then is ##f## injective?

- Relevant Equations
- ##f(X)=\{f(x):x\in X\}##

##f^{-1}(f(X))=\{x\in A:f(x)\in f(X)\}##

Injective: If ##f## is ##\textbf{injective}##, then given any two points ##y_1,y_2\in f(X)##, the pre-images of these points must be equal.

##A=(-1,1)##

##B=[0,1)##

Define ##f:A\longrightarrow B## by ##f(x)=x^2##

Set ##X=A##.

##f(X)=\{f(x):x\in X\}=\{x^2:x\in(-1,1)\}=B##

##f^{-1}(f(X))=\{x\in A:f(x)\in f(X)\}=\{x\in (-1,1):x^2\in B\}=X##

Now choose a non-zero point ##y\in f(X)##. There are two pre-images of this point: ##x,-x\in (-1,1)##. This implies that ##f## is not injective even though the inverse image of the image of ##X## via ##f## is equal to ##X##.

##B=[0,1)##

Define ##f:A\longrightarrow B## by ##f(x)=x^2##

Set ##X=A##.

##f(X)=\{f(x):x\in X\}=\{x^2:x\in(-1,1)\}=B##

##f^{-1}(f(X))=\{x\in A:f(x)\in f(X)\}=\{x\in (-1,1):x^2\in B\}=X##

Now choose a non-zero point ##y\in f(X)##. There are two pre-images of this point: ##x,-x\in (-1,1)##. This implies that ##f## is not injective even though the inverse image of the image of ##X## via ##f## is equal to ##X##.

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