3

Suppose $f$ is an injection. Show that $f^{-1}\circ f(x)=x$ for all $x\in D(f)$ and $f\circ f^{-1}(y)=y$ for all $y$ in $R(f)$.

In $f^{-1}$ it is defined as "Let $f$ be a one-one function with domain $D(f)$ in $A$ and range $R(f)$ in $B$. If $g=\{(b,a)\in B×A:(a,b)\in f\}$ then $g$ is a one-one function with domain $D(g)=R(f)$ in $B$ and with range $R(g)=D(f)$ in $A$. The function $g$ is called the function inverse to $f$."

I know that for it to be an injection the domain of $f$ maps into distinct elements of $y$, but where I am having trouble understanding and proving this is that, what does this $f^{-1}\circ f(x)=x$ mean? the $=x$ threw me off because what is $x$ in this case, could it have been an $x+3$ or $x^2$ instead of $x$?

Is proving this as easy as just showing that since it is injective then if $f(x_1)=f(x_2)$ $\implies$ $f^{-1}\circ f(x_1)=x_1=f^{-1}\circ f(x_2)=x_2$

Q.matin
  • 2,835
  • Are we assuming that $ \text{Dom}(f) \subseteq \mathbb{R} $? – Haskell Curry Feb 07 '13 at 07:23
  • @HaskellCurry Yes we are – Q.matin Feb 07 '13 at 07:24
  • If that is so, then $ (f^{-1} \circ f)(x + 3) = x + 3 $ is valid, as long as $ x + 3 \in \text{Dom}(f) $, or equivalently, $ x \in \text{Dom}(f) - 3 $. Also, $ (f^{-1} \circ f)(x^{2}) = x^{2} $ is valid as long as $ x^{2} \in \text{Dom}(f) $. – Haskell Curry Feb 07 '13 at 07:25
  • @HaskellCurry Agh, got it. I have edited it you think you can go over it and tell me if I am correct? – Q.matin Feb 07 '13 at 07:27
  • By definition, a function $ f: X \to Y $ is said to be injective if and only if $ f(x_{1}) = f(x_{2}) $ implies $ x_{1} = x_{2} $ for all $ x_{1},x_{2} \in X $. You can’t show that $ f $ is injective just by applying some $ f^{-1} $, because if you haven’t proven that $ f $ is injective in the first place, then how can you say that $ f^{-1}: \text{Range}(f) \to X $ exists? :) – Haskell Curry Feb 07 '13 at 07:33
  • @HaskellCurry But it says in the question it is injective so doesnt that mean i dont have to show it is injective? – Q.matin Feb 07 '13 at 07:43
  • Hmm... Well, I don’t understand why you need to introduce the two variables $ x_{1} $ and $ x_{2} $. To prove the first identity, note that if $ f(a) = b $, then $ {f^{-1}}(b) = a $. Let $ a = x $ and $ b = f(x) $, where $ x \in \text{Dom}(f) $. Clearly, $ f(a) = b $. Then $ {f^{-1}}(b) = a $ is just $ {f^{-1}}(f(x)) = x $, or equivalently, $ {f^{-1} \circ f}(x) = x $. The proof of the second identity is similar. – Haskell Curry Feb 07 '13 at 08:12
  • @Q.matin Are you sure you copied the question right? – Git Gud Feb 07 '13 at 18:53
  • @HaskellCurry thanks that makes it more clear! – Q.matin Feb 08 '13 at 00:57
  • @GitGud Yup, I just doubled check . – Q.matin Feb 08 '13 at 00:57
  • @Q.matin What's your definition of $f^{-1}$, given a function $f$? – Git Gud Feb 08 '13 at 06:56
  • @GitGud The def given in the book is "Let f be a one-one function with domain $D(f)$ in $A$ and range $R(f)$ in $B$. If $g={(b,a)\in B×A:(a,b)\in f}$ then $g$ is a one-one function with domain $D(g)=R(f)$ in $B$ and with range $R(g)=D(f)$ in A . The function $g$ is called the function inverse to $f$." – Q.matin Feb 08 '13 at 07:15
  • @Q.matin I suggest you add that information to the question. I gotta go now. When I get back in about 12 hours I'll help you out if no one else does in the meantime. – Git Gud Feb 08 '13 at 07:19
  • @GitGud That is fine, thanks a lot. Enjoy your 12 hour break . – Q.matin Feb 08 '13 at 07:26
  • 1
    @Q.matin You forgot to type { when you defined $g$ in the question.You just typed {. – Git Gud Feb 09 '13 at 01:15

1 Answers1

1

By definition $f^{-1}=\{(b,a)\in R(f)\times D(f) : (a,b)\in f\}$.

Now think of $f$ and $f^{-1}$ as binary relations and compute the composition of the relation $f^{-1}$ with the relation $f$. It comes $f^{-1}\circ f=\{(a,a')\in D(f)\times D(f):(\exists b\in R(f))\bigl( (a,b)\in f\wedge (b,a)\in f^{-1}\bigr)\}$.

(Recall that if $g$ is a function, then saying $g(z)=w$ is just shorthand notation for $(z,w)\in g$).

You want to prove that $(\forall x\in D(f))\bigl((f^{-1}\circ f)(x)=x\bigr)$, i.e., $(\forall x\in D(f))\bigl((x,x)\in f^{-1}\circ f\bigr)$.

Let $x\in D(f)$ be taken arbitrarily... (solution below).

Define $y=f(x)$. It follows that $(x,y)\in f$. Furthermore $(y,x)\in f^{-1}$. Since $y\in R(f)$ (by definition of $y$), it comes $(x,x)\in f^{-1}\circ f$, which means $(f^{-1}\circ f)(x)=x$.

The other one is similar.

Git Gud
  • 31,356