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Explain why $f^{-1}$ is a function if and only if $f$ is a bijective function.

My attempt:

$f^{1}$ is the inverse relation from B to A $\equiv$ function from B to A

By definition of a function from setA to setB, there is a relation from setA to B. (ARB? relation) such that is satisfies two properties:

1) All element in the domain A must be related to some function in the codomain B

2) No element in A is related to more than one element in B.

Hence, that would mean that 1) the function must be onto and 2) the function must be one-to-one.

Since the function from A to B has to be bijective, the inverse function must be bijective too.


It's hard for me explain. But basically because the function from A to B is described to have a relation from A to B and that the inverse has a relation from B to A. Since the relation from A to B is bijective, hence the inverse must be bijective too. Also, does bijection has anything to do with equivalence relation on a single set where a relation R on set A must be reflexive, symmetric and transitive.

Thanks!

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I recommend you explain it like this:

Let $f : A \to B$ be a bijection. Let $f^{-1}$ be the inverse function of $f$.

We want to show $f^{-1}$ is a bijection, so we need to show $f^{-1}$ is one-to-one and onto.

First, let's show $f^{-1}$ is onto:

Since $f^{-1} : B \to A$, we need to show for each $a \in A$, there is some $b \in B$ so that $f^{-1}(b) = a$. Well, let $a$ be any element of $A$, then. Since $f : A \to B$, $f(a)$ is an element of $B$. But by definition of the inverse function $f^{-1}$, this map sends $f(a)$ from $B$ to $a$ in $A$. So that means we found a $b \in B$ with $f^{-1}(b) = a$, namely, the element $b = f(a)$.

Now to show $f^{-1}$ is one-to-one:

Since $f^{-1} : B \to A$, we need to show if for any $b_{1}, b_{2} \in B$, we have $f^{-1}(b_{1}) = f^{-1}(b_{2})$, then it should hold that $b_{1} = b_{2}$. Well then, suppose $f^{-1}(b_{1}) = f^{-1}(b_{2})$. Since $f : A \to B$ is onto, we can find $a_{1}$ and $a_{2}$ in $A$ so that $f(a_{1}) = b_{1}$ and $f(a_{2}) = b_{2}$. So we have $f^{-1}(b_{1}) = f^{-1}(b_{2})$ implies $f^{-1}(f(a_{1})) = f^{-1}(f(a_{2}))$. But by definition of the inverse function, $f^{-1}$ sends the element $f(a)$ to $a$, so $f^{-1}(f(a_{1})) = a_{1}$ and $f^{-1}(f(a_{2})) = a_{2}$, and so this implies $a_{1} = a_{2}$. Since $a_{1} = a_{2}$, and $f$ is a function (and thus well-defined), it can't send one element to two different elements, so that means $f(a_{1}) = f(a_{2})$, i.e., $b_{1} = b_{2}$.

Note: where did we use the fact that the original function $f$ is one-to-one and onto here? Well, we didn't use one-to-one directly in the arguments above (but onto was used in the second argument). But they are still necessary. If $f : A \to B$ is onto, then that would allow us to construct a potential inverse with domain $B$. If $f$ were not onto, our inverse function could not have domain $B$. Similarly, if $f$ is one-to-one, then that allows us to construct a well-defined inverse, i.e., an inverse that's actually a function (i.e., it doesn't send one element to more than one element). Both of these are necessary to discuss the existence of a function $f^{-1}: B \to A$ defined in the usual way an inverse is defined.

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    $\begingroup$ @misheekoh Yes, you can definitely have a function that is one-to-one but not onto and it's still a function. You can also have a function that's onto but not one-to-one and it's still a function. $\endgroup$ – layman Dec 6 '15 at 0:41
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    $\begingroup$ @misheekoh A function $f : A \to B$ is just a rule that sends stuff from $A$ to stuff in $B$. And every function must be well-defined. That just means no single element from $A$ can be sent to two elements in $B$. So for example $f : \Bbb R \to \Bbb R$ which sends the element $1$ to both $0$ and $3$ (i.e., $f(1) = 0$ and $f(1) =3$) is not a function. $\endgroup$ – layman Dec 6 '15 at 0:42
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    $\begingroup$ @misheekoh Incorrect! Well defined and injective look similar but are very different. Well defined means no input can correspond to more than one output. One-to-one means no output can have more than one input being sent to it. $\endgroup$ – layman Dec 6 '15 at 2:45
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    $\begingroup$ @misheekoh Make sure you understand my last comment. It's a common misconception. Every function by definition must be well defined. No input can correspond to more than one output. But functions don't have to be one-to-one. A function can have an output that has more than one input. Take $f : \Bbb R \to \Bbb R$ defined by $f(x) = 2$. It's the constant function $2$. This is a function, since it is well-defined. No input corresponds to more than one output since every input has the output $2$. But $f$ is not one-to-one. There is an output with more than one input, namely... $\endgroup$ – layman Dec 6 '15 at 2:46
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    $\begingroup$ @misheekoh ...the output $2$. Every input is sent to the output $2$, so $f$ is not one-to-one. $\endgroup$ – layman Dec 6 '15 at 2:47
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If $f:A \rightarrow B$ is a function then it must be the case that $\forall a \in A, f(a)$ is unique. If $f^{-1}: B \rightarrow A$ exists, then it must be the case that $\forall b \in B, f^{-1}(b)$ is unique. So, each element in A is paired with one element in $B$ (and vice versa), specifically, $a \in A$ is paired with $f(a) \in B$.

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I assume your $f^1$ notation actually means $f^{-1}$.The function $f$ has to be bijective because otherwise, our so called "$f^{-1}$" would not even be a function. Let's see why:

1) Assume $f:A\rightarrow B$ is not surjective: then there is an element $b' \in B$ that doesn't have a preimage. Therefore if we attempt to construct a relation from every $b \in B$ to an $a\in A$, so that $b\mathcal Ra$ if $ b=f(a)$, $b'$ is left behind, then $\mathcal R$ cannot be a function.

2)Assume $f:A\rightarrow B$ is not injective. Then for some $a_1 $ and $ a_2$ in $A$ we have that $f(a_1)=f(a_2)$, then if we attempt to construct the same relation $\mathcal R$ we conclude it cannot be a function because for some $b \in B$ we have that $b\mathcal Ra_1$ and $b\mathcal Ra_2$

Your explanation works, I just tried to explain it in a more technical way. And no, $f$ being an equivalence relation has nothing to do with $f^{-1}$ existing

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