## Bijection

Let $$f\colon X \to Y$$ and $$g\colon Y \to X$$ be functions such that for every $$x \in X$$, $$(g \circ f)(x) = x$$ and for every $$y \in Y$$, $$(f \circ g)(y) = y$$. Prove that $$f$$ and $$g$$ are bijective (i.e. both injective and surjective).

• #### Hint

Try to show that if $$g \circ f$$ is injective, then $$f$$ is also injective, and also that if $$f \circ g$$ is surjective, then $$f$$ is also surjective.

• #### Solution

By contradiction, if $$f$$ is not injective, there exist $$x$$ and $$y$$, $$f(x)=f(y)$$, but then $$(g\circ f)(x)=(g\circ f)(y)$$, but this cannot equal $$x$$ and $$y$$ at the same time.

Similarly, if $$f$$ is not surjective, there exists $$y\in Y$$ which is not the image of any element in $$X$$, but then $$(f \circ g)(y)\ne y$$.

So $$f$$ is bijective.

We can show this for $$g$$ by the same argument, exchanging $$X$$ and $$Y$$ and the order of $$f$$ and $$g$$.

Note: If functions $$f$$ and $$g$$ fulfill the given conditions, then they must be inverse functions of each other.