 #1
Eclair_de_XII
 1,066
 90
 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 preimages 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 nonzero point ##y\in f(X)##. There are two preimages 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 nonzero point ##y\in f(X)##. There are two preimages 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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