Bijective mappings and transpositions While self-teaching bijective mappings I came across the following question: 
Given that the bijective mapping $f : X \mapsto Y$ exists and $x$ and $y$ are elements of $X$ and $Y$ respectively, prove that there exists a bijection $g : X \mapsto Y$ such that $g(x)=y$. 
I am still new to the concepts of bijective mappings etc. but the proof I came up with was as follows:
If $y = f(x)$ then it is obvious that such a function $g$ exists and that $g=f$, and since $f$ is bijective then $g$ must also be bijective. Now suppose that $f(x) = y_0 \neq y$. Then there will exist a composite function $g(x) = (\tau_{y_0, y}\circ f)(x) = y$ where the $\tau$ notation is a transposition of elements and is bijective, so then there must exist some $g(x) = y$ 
Am I going about the proof in the wrong way, or is this OK?
 A: Your proof is correct and is a great way to approach it.  To make it a bit clearer, you should mention that you are defining $g=\tau_{y_0,y}\circ f$ (as you wrote it it looks like you are asserting that they agree only when evaluated at $x$) and that $g$ is a bijection because a composition of bijections is a bijection.  The phrase "then there must exist some $g(x) = y$" is also not quite right: what you want to say is "there must exist some bijection $g:X\to Y$ such that $g(x)=y$".
A: I think what you are saying is right but the way it is written means it is a but unclear what you mean.
More explicitly you could show the result as follows:
$f:X\rightarrow Y$ is bijective and so $f^{-1}:Y\rightarrow X$ is well defined and bijective also. Then let $y_0 = f(x)$ and $x_0 = f^{-1}(y)$. Now define the function $g$,
$$
g:X\rightarrow Y
$$
where $g(t) = f(t)$ for all $t \neq x,x_0$ and then $g(x) = y$, $g(x_0) = y_0$. You should easily be able to show that $g$ is bijective and it clearly satisifes the given condition.
