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I know that for a function $f$ there exists an inverse $f^{-1}$ when $f$ is one-one and onto in its domain. I also know that a function $f$ and its inverse $f^{-1}$ are mirror images about the line $y=x$.

Now, can we say that when two functions which are exactly mirror images about the line $y=x$, are inverses of each other? Or in other words is the converse of the statement "Function and its inverse are mirror images of each other about the line $y=x$" is always true? If it is not always true, kindly give me circumstances when the converse fails.

Edit:

From this Quora answer, it is said that two functions having the same graph need not necessarily be equal. Then how can we conclude that the graph mirror imaged about the line $y=x$ is definitely its inverse?

Vishnu
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2 Answers2

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Yes, because if one function is $y = f(x)$, the other is $x = g(y)$, since swapping $x$ and $y$ has the same effect as reflecting across the line $y=x$.

Substituting the first into the second, $x = g(y) = g(f(x))$, or in other words, $g^{-1} (x) = f(x)$ if $g^{-1} (x)$ exists. Similarly, substituting the other way around, $y = f(x) = f(g(y)$, so $f^{-1} (y) = g(y)$ if $f^{-1} (y)$ exists.

Toby Mak
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  • I might make them different functions.. This is making it seem like $f$ is an involution which it need not be. – Cameron Williams Oct 07 '19 at 11:48
  • Thanks for the comment. I have now edited. – Toby Mak Oct 07 '19 at 11:51
  • Thanks for your answer. I could not think of any situation when the converse doesn't hold good. So, are there any circumstances when two functions are mirror images about $y=x$ but not inverses of each other? – Vishnu Oct 08 '19 at 06:03
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    No, because if two functions are inverses of each other, then there is a bijection between the two functions. For example, if $f(x) = x+2$, $f^{-1} (x) = x-2$, and the point $(7,9)$ of $f(x)$ corresponds to $(9,7)$ on $f^{-1} (x)$. Since the mirror image exists, then there has to be a bijection where you can construct an inverse. – Toby Mak Oct 08 '19 at 06:17
  • @TobyMak, Could you please read the question again, I have added some details. – Vishnu Oct 09 '19 at 09:51
  • You are scaring yourself here. Rewriting $(y-x)(x^2+y^2)$ to isolate $y$, you get $y=x$, and rewriting $y-x=0$ you also get $y=x$. This shows that for both functions, each value of $x$ still corresponds to the same value of $y$. – Toby Mak Oct 09 '19 at 09:57
  • I suppose we should add in that both functions can be continuous, since you cannot really talk about bijections unless there is a value of $x$ for each value of $y$. Also, both functions should be one-to-one, unless you restrict them to a particular range which makes them one-to-one (such as $y=x^2$ in the range $x ≥ 0$). – Toby Mak Oct 09 '19 at 09:59
  • @TobyMak, WRT your first comment today, I understand that. But still the formulae for the two cases look different - for one is of order 3 and the other of order 1. – Vishnu Oct 09 '19 at 10:03
  • So two functions with same graphs are necessarily equal, and not unequal as said in the Quora answer. – Vishnu Oct 09 '19 at 10:03
  • As far as I know, two functions are equal if and only if they have the same formula and same domain. Are there any exceptions to this rule? – Vishnu Oct 09 '19 at 10:10
  • Let us continue this discussion in chat. – Vishnu Oct 09 '19 at 10:19
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Suppose the two functions whose graphs are reflections over the line $y=x$ are named $f$ and $g$. What does it even mean to say the graphs are reflections over the line $y=x$? It means that if $(a,b)$ is some point on $f$'s graph, then $(b,a)$ is a point on $g$'s graph.

So take any $a$ in $f$'s domain, and let $b=f(a)$. Then the point $(a,b)$ is on $f$'s graph. So $(b,a)$ is on $g$'s graph. So $g(b)=a$. So $g(f(a))=a$. And this was for an arbitrary $a$ is $f$'s domain. The argument is symmetric for showing that for any $b$ in $g$'s domain, that $f(g(b))=b$. So the conclusion is yes, $f$ and $g$ are inverses.

2'5 9'2
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  • Thank you for your answer. In this Quora answer, it is given that two functions having same graph need not be equivalent and here at math.se it totally confuses me. I completely understood your answer. But based on that Quora answer, it is also equally correct to claim the graph mirrored about $y=x$ may not be the inverse of the original graph. It would be great if you could explain this. – Vishnu Oct 18 '19 at 06:29
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    @Intellex Do you understand the difference between a "function" and an "equation"? That Quora answer is about two equations in $x$ and $y$, and the solutions to the two equations make the same graph. But two functions with the same graph are indeed the same function. – 2'5 9'2 Oct 18 '19 at 15:13