Question
Consider the function g: (x,y) (x^2,y^2). Assume that the domain of g is the entire xy-coordinateplane. a) What is the range of g? b) Is the function g a one-to-one function? Explain. c) Is the function g an onto function? Explain. d) Does the fuction g have any fixed points? Is so,identify them. If not, explain how you know there aren'tany. I'm still just struggling with the ideas of injective andsujective functions. I believe this function is surjectivebut not injective, but I'm really not sure. It seems that anyx in the domain will easily map to a y in the codomain. And Iwould think that considering the negative numbers, there will bemultiple outputs coming from the same input. As for whetherit has any fixed points, I have no idea. Any help would begreatly appreciated. Thanks. Consider the function g: (x,y) (x^2,y^2). Assume that the domain of g is the entire xy-coordinateplane. a) What is the range of g? b) Is the function g a one-to-one function? Explain. c) Is the function g an onto function? Explain. d) Does the fuction g have any fixed points? Is so,identify them. If not, explain how you know there aren'tany. I'm still just struggling with the ideas of injective andsujective functions. I believe this function is surjectivebut not injective, but I'm really not sure. It seems that anyx in the domain will easily map to a y in the codomain. And Iwould think that considering the negative numbers, there will bemultiple outputs coming from the same input. As for whetherit has any fixed points, I have no idea. Any help would begreatly appreciated. Thanks.
Explanation / Answer
Assuming that the codomain of this function is the set of realnumbers R: b) g is not one-to-one because (except in the case wherex=y=0) there is more than one argument (x,y) that maps to eachoutput (x^2, y^2). c) g is not onto because there is no argument (x,y) such thatx^2=-c for some scalar c and similarly for y^2. d) g has fixed points, which are defined as the points wherethe function argument equals the output, i.e., g(x,y)=(x,y). Inthis case, the set of fixed points fulfill the conditions thatx=x^2 and y=y^2. So x={0, 1} and y={0, 1}. Thus, the set of fixedpoints are (x,y): {(0, 0), (0, 1), (1, 0), (1,1)}.