Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $f: X \to Y$ such that $f$ is a bijection prove the existence of a $g:Y\to X$ such that:

$f \circ g = 1_Y $


$g \circ f = 1_X $

Now since $f$ is bijective $\forall y \in Y: \exists!x \in X $ s.t $f(x)= y$

If I define $g(f(x)) = x = 1_X(x) \ \forall f(x) \in Y $

Then for $x \in X $ I notice that:

$g \circ f (x) = g(f(x)) = 1_X(x)$

I am have some trouble proving the remaining part. How would I continue from here?

share|cite|improve this question
up vote 1 down vote accepted

To define $g : Y \rightarrow X$: For each $y \in Y$, since $f$ is surjective and injective, there exists a unique $x_y$ such that $f(x_y) = y$. Let $g(y) = x_y$.

Observe that $x_{f(x)} = x$.

For all $x \in X$, $g(f(x)) = x_{f(x)} = x$. Hence $g \circ f = \text{id}_X$

For all $y \in Y$, $f(g(y)) = f(x_y) = y$. Hence $f \circ g = \text{id}_Y$

share|cite|improve this answer
I thought as much. – Mathman May 6 '14 at 9:33

notice that $g(y)$ is exactly defined as the value (unique, by injectivity of $f$) such that $f(g(y))=y$.

share|cite|improve this answer
I get g(y) has a unique value, but how exactly is one able to say $f(g(y))=y$ – Mathman May 6 '14 at 9:29
fix $y$. Then you get $x$ such that $y=f(x)$. Then $y=f(x)=f(g(f(x)))=fg(y)$. – Franco May 6 '14 at 10:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.