Let $f : A \rightarrow B$ be an arbitrary function.

Question

(a) Prove that if $f$ is a bijection (and hence invertible), then $f^{−1}(f(x))=x \ \forall x \in A$, and $f(f^{−1}(x))=x \:\:\forall\:\: x \in B$.

(b) Conversely, show that if there is a function $g : B \rightarrow A$, satisfying $g(f(x)) = x \:\:\forall \:\: x \in A$, and $f(g(x)) = x \:\: \forall \:\:x \in B$ then $f$ is a bijection, and $f^{−1} = g$.

Apologizes for the poor formatting.

This makes sense logically, the composition of $f$ and $f^{-1}$ is always going to be the input but I am not sure how to formally prove this .

To my knowledge yes it is. This is a confusing question because it looks straightforward but I am unsure how to formally prove this — smith sam, Nov 21 '18 at 18:34
I vote to close this question because it is missing details. For instance what is the definition for $f^{-1}$ that you're given and what did you try? Right now I don't think anyone can "solve" this question. — Yanko, Nov 21 '18 at 18:36
The definition that I am given is "the inverse of f is a function g:B-->A that assigns to any y∈B, the only x∈A for which f(x)=y". Sorry I am new and am unfamiliar on how to use this website. — smith sam, Nov 21 '18 at 18:41
It's close to the definition. But it's not entirely. I'd say. $f^{-1}(w)$ is by definition the unique $z$ so that $f(z) =w$. So $f^{-1}(f(x))$ is the unique $z$ so that $f(z) = f(x)$. And $f$ is one-to-one $f(z) =f(x)$ means $z = x$ so the unique $z$ is $x$. — fleablood, Nov 21 '18 at 18:44
You can have a look at this post: Inverse of a Function exists iff Function is bijective (and other questions linked there). — Martin Sleziak, Nov 25 '18 at 12:48

score 1 · Answer 1 · answered Nov 21 '18 at 18:56

a) is almost a definition.

$f:A\to B$ is a function. That means for every $x \in A$ then there is a $w_x \in B$ so that $f(x) = w_x$. That's .... not anything deep. That's just what a function is. (But my notation may seem strange to you.)

$f$ is a bijection so it is invertible. That means by definition that for any $y\in B$ that there is exactly one $x_y \in A$ so that $f(x_y) =y$.

So for any $x \in A$ there is a unique $w \in B$ so that $f(x) = w$. And for $w \in B$ there is a unique $z= f^{-1}(w)\in A$ so that $f(z) = w = f(x)$. But if that value is unique and $f(x)$ is also equal to $w$ that means $z=f^{-1}(w) = x$.

So $f^{-1}(f(x)) = f^{-1}(w) = x$.

Shubham Johri · Answer 2 · 2018-11-23T11:04:19.463

To make sense of the question, note that for $f: A \to B$, $f^{-1}(P)$ represents the inverse image of a set $P\subseteq B$ under $f$. If $f(x)$ is treated as the set $\{f(x)\}$, then, by definition of inverse image,

$$f^{-1}(\{f(x)\}) = \{\ a : f(a)=f(x),\ a \in A \}$$

Since $f$ is a bijection, it is injective. Therefore,

$$f(a)=f(x)\implies a=x\\ \implies f^{-1}(\{f(x)\})=\{x\}, \forall\ x \in A\\$$

You can prove that $f(f^{-1}(x))=x, \ \forall\ x \in B$, similarly.

For the second part, observe that $g(f(x))=x, \forall\ x \in A,$ is a bijective function. $g(f(x))$ is injective $\implies f$ is injective and $g(f(x))$ is surjective $\implies g$ is surjective. Similarly, $f(g(x))=x, \forall\ x \in B,$ is a bijective function. $f(g(x))$ is injective $\implies g$ is injective and $f(g(x))$ is surjective $\implies f$ is surjective.

Thus, if such a $g$ exists, then both $f$ and $g$ are bijective. Let the inverse of $f$ be denoted by $f^{-1}$, $$f(g(x))=x, \forall\ x \in B\\ \implies f^{-1}(f(g(x))=f^{-1}(x),\ \forall\ x \in B\\ \implies g(x)=f^{-1}(x),\ \forall\ x \in B$$

Let $f : A \rightarrow B$ be an arbitrary function.

2 Answers2