Confused about one to one functions and cardinality

Question

There is something that I'm not getting about functions and cardinality of sets.

I've read the following:

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If $F$ is a one to one function, then $F^{-1}$ (the inverse) is also a one-to-one function.

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Because when I read about cardinality of sets, I encounter with the definition:

$|A|\leq |\mathcal P(A)|$ if there is a one to one function from $A$ to $\mathcal P(A)$.

This function is trivial $F(a) = \{a\}$ and it's one to one and not onto.

But if I follow the statement from above of $F^{-1}$ (the inverse of $F$) being also one to one, this leads me to the following :

$|\mathcal P(A)|\leq |A|$

Which I know is false. So there is obviously something wrong with my understanding, what concept am I missing, what I'm getting wrong?

Answer 1

While it is correct that if $F$ is an injective function, so is $F^{-1}$, you forget to consider what is the domain of $F^{-1}$.

Recall that when writing $F\colon X\to Y$, we mean that $F$ is a function whose domain is all of $X$, and its range is a subset of $Y$. The inverse function is from the range of $F$, which may or may not be all of $Y$. And indeed in the case of $F(a)=\{a\}$, the range of $F$ is only a small part of $\mathcal P(A)$.

Answer 2

Let $F$ be a function from a domain $X$ to a codomain $Y$.

$F^{-1}$ need not exist solely because of $F$ being one-to-one. You will have $F^{-1}|_{range(F)}$ will exist and will be one-to-one, however since $F$ is not guaranteed to hit every element in the codomain, you are not guaranteed that $F^{-1}(y)$ is even defined for each $y\in Y$.

The missing bit of information needed to require $F^{-1}$ to exist is that $F$ be also onto. If $F$ is both one-to-one and onto (i.e. is bijective), then the inverse exists and is also bijective.