If $f: A \rightarrow B$ is injective, $\exists ! f^{-1}:f(A)\rightarrow A$ such that $f^{-1}f(x) = x$

I know the proof of this is super simple, like 2 lines (+ some for uniqueness), but I can't figure how to logically work it out. For definition of injective function, I have $\forall x,y\in A, f(x)=f(y) \implies x=y$.

So we are trying to show that for all $y \in f(A),$ there exists a unique $x\in A$ with $f^{-1}(y) = x$. Could I change y to f(x) right there?

  • $\begingroup$ In your question doesn't " There exists ! " mean : "There does not exist"? Should the ! not be there? $\endgroup$ – Adam Rubinson Nov 28 '12 at 13:44
  • $\begingroup$ It means "there exists a unique". If it were "not", it would go before the existential quantifier, but that is CS notation anyways. To say "there does not exist" would be $\not \exists$ or $\neg \exists$. See en.wikipedia.org/wiki/Uniqueness_quantification $\endgroup$ – fhyve Nov 29 '12 at 2:18
  • $\begingroup$ Ah right, thanks fhyve! I didn't know that! In that case everything seems correct. $\endgroup$ – Adam Rubinson Nov 29 '12 at 9:19

Note that the injectivity is needed for uniqueness only: If you are given $y \in f(A)$, then by the definition of $f(A) = \{f(x)\mid x \in A\}$ there exists a $x \in A$ with $f(x) = y$.

For uniqueness use injectivity: If $f(x) = y$ and $f(x') = y$, then $f(x) = f(x')$, hence $x=x'$.

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  • $\begingroup$ Really? I was sure that the injectivity was needed to define the function as well... $\endgroup$ – Asaf Karagila Nov 26 '12 at 8:04
  • $\begingroup$ @AsafKaragila I was reffering to the OP last paragraph. It is written there "we are trying to show that for all $y∈f(A)$, there exists a unique $x \in A$ [...]". Of course for having defined a function by this, we need injectivity. $\endgroup$ – martini Nov 26 '12 at 8:08
  • $\begingroup$ I should probably wake up completely before making further remarks. :-) $\endgroup$ – Asaf Karagila Nov 26 '12 at 8:09
  • $\begingroup$ @AsafKaragila Want some coffee? ;-). $\endgroup$ – martini Nov 26 '12 at 8:12

We can use the following:

Lemma 1: A function $f : X \to Y$ is invertible if and only if it is one-to-one and onto.

Proof: Suppose that $f : X \to Y$ is invertible. Then, there exists a function $g : Y \to X$ such that $(f \circ g)(x) = (g \circ f)(x) = x$. To see that $f$ is onto, pick an element $y \in Y$. Then, $f(g(y)) = y$ so that $y$ is in the range of $f$. To see that $f$ is one-to-one, suppose that $f(x) = f(y)$. Then, $$g(f(x)) = g(f(y)) \implies x = y.$$

Now, suppose that $f$ is one-to-one and onto. We will show that $f$ is invertible. As $f$ is onto, we have that $\forall y \in Y, \exists x \in X$ such that $y = f(x)$. Since $f$ is injective, we have that $f(x) = f(y) \implies x = y, \forall x , y \in X$.

Let $g : Y \to X$ be defined by $g(y) = x$ if and only if $f(x) = y ~$ (why can we assume such a $g$ exists?). Then $g(f(x)) = g(y) = x$ and $f(g(y)) = f(x) = y$. Hence $g$ is an inverse of $f$. QED.

Lemma 2: The inverse of a function $f$ is unique if it exists.

Proof: Suppose that $f : X \to Y$ is invertible and it has two inverses $g, h : Y \to X$. Note that $\forall x \in X$, we have $f(g(x)) = x = f(h(x))$. Hence: $$ g(x) = g(f(g(x))) = g (f(h(x))) \implies g(x) = h(x). $$ Since $x$ was arbitrary, $g$ and $h$ must be identical as they have the same value for all $x \in X$. QED.

Finally, apply the above lemmas to the function $f\big|_A : A \to f(A)q$. Note that the function $f : A\to f(A)$ is bijective so long as it is injective. By definition, $$ f(A) = \left\{y \in B : f(x) = y ~~~\text{ for some }~~ ~x \in X\right\}, $$ so that $f\big|_A : A \to f(A)$ is surjective, and hence $f|_A$ is bijective.

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