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We have this question for homework:

Let $h: B\to A$ be the left inverse of $f$, and $g:B\to A$ the right inverse of $f$. Prove that that $h$ and $g$ are the same function.

To prove this I stated "Let $a\in A$ and $b\in B$ such that $f(a)=b$ and $g(b)=a$."

My question is: is that even legal to assume so?

If it is then all there's left to show is that $h(f(a)) = h(b) \neq g(b) = a$ is a contradiction to $h$ being an inverse function of $f$...

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It might be best not to even deal with specific elements of $A$ and $B$. I assume $f\colon A\to B$. You're given that $h$ is the left inverse and $g$ the right inverse, so $h\circ f=\text{id}_A$ and $f\circ g=\text{id}_B$. Then $ h=h\circ\text{id}_B=h\circ(f\circ g)=\cdots $ and remember that composition of functions is associative.

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let g:B→A and h:B→A be two inverse of function f:A→B the we are to proof that g=h solution

let b∈B such that g(b)=p and h(b)=q since g and h are inverse functions these imply that f(p)=b and f(q)=b therefore f(p) = f(q) but f is one to one these imply that p=q g(b) = h(b) since b is arbitrary vb∈B we say g=h hence proved!