Testing for a function

[fresh_42]

Is it that $\displaystyle y = x^2$ has no inverse or that it has an inverse which is not a function?

The inverse being $\displaystyle x = \pm \sqrt y$

That depends on how you define an inverse. $y=x^2$ has no inverse function and we can say it has no inverse.

An inverse is a unary operation. It requires to know on which set!

If we use the language (set) of relations then inverting a relation is simply interchanging the components. If there is a relation $R$ between the sets $A$ and $B$, sometimes written $ARB$ or $A\sim_R B$ then $R$ is a subset of $A\times B .$ Say we have $R=A\times B$ to keep it simple. This means $R=\{(x,y)\,|\,x\in A, y\in B\} .$ The inverse relation would then be $R^{-1}=\{(y,x)\,|\,x\in A,y\in B\} .$ Since $f$ is a relation, we can define a relation $f^{-1}$ by $f^{-1}=\{(f(x),x)\}$ instead of $f=\{(x,f(x))\} .$

If $f(x)=x^2$ then $(-1,1),(1,1)\in f$ and $(1,-1),(1,1) \in f^{-1} .$ However, this isn't a function any longer, since this would mean that we have $f^{-1}(1)=-1$ and $f^{-1}(1)=1 .$

Long story short ...

The inverse being $f^{-1}(y)= x=\pm \sqrt{y}$

... which is no function.

 

[Agent Smith]

That depends on how you define an inverse. $y=x^2$ has no inverse function and we can say it has no inverse.

An inverse is a unary operation. It requires to know on which set!

If we use the language (set) of relations then inverting a relation is simply interchanging the components. If there is a relation $R$ between the sets $A$ and $B$, sometimes written $ARB$ or $A\sim_R B$ then $R$ is a subset of $A\times B .$ Say we have $R=A\times B$ to keep it simple. This means $R=\{(x,y)\,|\,x\in A, y\in B\} .$ The inverse relation would then be $R^{-1}=\{(y,x)\,|\,x\in A,y\in B\} .$ Since $f$ is a relation, we can define a relation $f^{-1}$ by $f^{-1}=\{(f(x),x)\}$ instead of $f=\{(x,f(x))\} .$

If $f(x)=x^2$ then $(-1,1),(1,1)\in f$ and $(1,-1),(1,1) \in f^{-1} .$ However, this isn't a function any longer, since this would mean that we have $f^{-1}(1)=-1$ and $f^{-1}(1)=1 .$

Long story short ...

... which is no function.

There's a lot to learn from your posts, but the process - looking for a needle in a haystack. Didn't know mathematicians could be so verbose.