# Proving a function is bijective

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1. Nov 1, 2014

### schlynn

Mod note: Moved from a technical section, so missing the homework template.
Here is what I'm trying to prove.

Let f:A->B. If there are two functions g:B->A and h:B->A such that g(f(a))=a for every a in A and f(h(b))=b for every b in B, then f is bijective and g=h=f^(-1).

I think I have most the proof, I started by showing g(f(a))=a implies g o f(a)=a and that maps A->B->A all exactly once, and a similar argument shows f(h(b))=b implies that the mapping is B->A->B all exactly once again, and this seems like the definition of an inverse, so there for it's bijective? Is that the right way to approach this proof?

Last edited by a moderator: Nov 1, 2014
2. Nov 1, 2014

### gopher_p

OK, to help you see where your intuition is leading you astray, let $f:\mathbb{R}^{\geq0}\rightarrow\mathbb{R}$ be given by $f(x)=\sqrt{x}$ and $g:\mathbb{R}\rightarrow\mathbb{R}^{\geq0}$ by $g(x)=x^2$. Clearly $g(f(a))=a$ for all $a\in\mathbb{R}^{\geq0}$, but $f$ is not bijective.

The point of the exercise is to show that the existence of $g$ and $h$ in tandem give that $f$ is a bijection, which then leads to $g$ and $h$ being inverses.

3. Nov 1, 2014

### schlynn

OK, I see where I went wrong, but how do I use the existence of g and h to show that the function is a bijection? It has to be injective and surjective, I know the definition of them but don't see how g and h show it's bijective. Can you point me in the right direction? Does 1 function show one property and the other function the other property? These types of proofs are new to me.

4. Nov 1, 2014

### gopher_p

Yes.

If $f(a_1)=f(a_2)$, then $g(f(a_1))=g(f(a_2))$, right?

Given $b\in B$, you know that $f(h(b))=b$, right?

I think that's the extent to which I can point you in the right direction.

5. Nov 2, 2014

### schlynn

Ok, I think I understand this now, so the first part shows injectivity and the second shows surjectivity, and the order of the composition changes to show that the sets your mapping actually allow for an inverse, and then bijectivity follows?