If two functions are mirror images of each other about the line $y=x$, are they inverses of each other? 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?
 A: 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.
A: 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.
