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Let f : Z → Z be defined by f(x) = x^2 + 1, and let C = {1, 2, 3}

So f doesn't have an inverse but can I still do such operations:

f^-1(f(C)) = C and f(f^-1(C)) = C

In addition, would f^-1(C) equal the empty set? Or can I not even assume f^-1 equals anything. I appreciate any help, I'm curious in this matter.

Tingo Hugo
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    Compare the phrase "inverse function" and "inverse relation." Your $f$ does have an inverse relation but not an inverse function. – JMoravitz Dec 06 '22 at 18:40
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    As for $f^{-1}(C)$, the question is whether there exist any values of $x$ such that $x^2+1$ is equal to one of ${1,2,3}$ and the answer is that yes. There are three such numbers. As for whether $f^{-1}(f(C))=f(f^{-1}(C))=C$ or not, the answer is no, they are not equal. – JMoravitz Dec 06 '22 at 18:42
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    The notation $f^{-1}$ has two different meanings which you're mixing up with each other. For real numbers $x$, $f^{-1}(x)$ is an inverse function which might or might not be defined depending on the function $f$, but when it is defined it satisfies $f(f^{-1}(x))=x$ and $f^{-1}(f(x))=x$. For sets $C$, $f^{-1}(C)$ is the inverse image ${y\colon f(y)\in C}$; it is always defined for every function $f$ but does not automatically satisfy $f(f^{-1}(C))=C$ or $f^{-1}(f(C))=C$. – Greg Martin Dec 06 '22 at 18:42

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Your $f$ is not bijective and so it does not have an inverse in terms of inverse functions. In terms of relations and inverse relations, we have $f(C)$ with $C$ a set is defined as $f(C):=\{f(x)~:~x\in C\}$, that is to say it is the set of images of elements of $C$.

Here, $f(\{1,2,3\}) = \{1^2+1,2^2+1,3^2+1\}=\{2,5,10\}$

$f^{-1}(C)$ with $C$ a set is defined as $f^{-1}(C)=\{x~:~\exists c\in C~f(x)=c\}$, that is to say, it is the set of preimages of elements of $C$.

Here $f^{-1}(C) = \{-1,0,1\}$ since $(-1)^2+1=2$ is an element of $C$, as is $0^2+1=1$ and $1^2+1=2$.

Note that $f^{-1}(f(C)) = f^{-1}(\{2,5,10\})=\{-3,-2,-1,1,2,3\}$ and that $f(f^{-1}(C))=f(\{-1,0,1\})=\{1,2\}$. Neither of which are equal to $C$.

Now... in the event that $f$ actually was bijective, we would have had $f(f^{-1}(C))=C$ but that is not the case here hence why we find them to be unequal.

JMoravitz
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