Htykuut

To prove that $f$ and $g$ are mutually inverse bijections $Ato B$ and $Bto A$, it is necessary to prove:

• that really $f:Ato B$ and $g:Bto A$;


• for every $x_0in A$ and $y=f(x_0)$ we have $x_1=x_0$ for $x_1=g(y)$;


• for every $y_0in B$ and $x=g(y_0)$ we have $y_1=y_0$ for $y_1=f(x)$.

Now suppose we already know that $f$ is an injection. Knowing this which items of the above list may be skipped without breaking the proof that they are mutually inverse?

It's a bit easier if we rephrase your conditions thus:

• 
  $f in A to B$ and $g in B to A$
  


• for all $a in A$, $g(f(a)) = a$
  


• for all $b in B$, $f(g(b)) = b$
  

If the third condition holds, and we put $f(a)$ for $b$, then we get:
$$
f(g(f(a)) = f(a)
$$
If also $f$ is an injection (so that $f(x) = f(y)$ implies $x = y$), then we can "cancel" the $f$ on the two sides of the above equation giving:
$$
g(f(a)) = a
$$
which is the second condition.

So if $f$ is an injection you don't need to check the second condition. (By a somewhat similar argument, if $g$ is a surjection, you don't need to check the third condition.)