Proof on Endomorphism

Question

Let A be element of Endomorphism of V (V is a finite dimensional vector space over F) such that A is onto. Assume that there exist a function B: V $\to$ V such that BA = I. Prove that AB = I

Can you give me a hint on how to prove this problem? Thanks.

Here is working solution. Since A is onto, there exist x in V such that A(x) = v.

We need to show that BA = I.

(BA)(x) = B(A(x)) = B(v) then I don't know what's next

There is no need for $V$ being a vector space. This is valid for surjective functions $f: X \to Y$. — André Caldas, Oct 31 '12 at 18:02
See http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i, http://math.stackexchange.com/questions/216569/assuming-ab-i-prove-ba-i, etc. — wj32, Nov 01 '12 at 08:39

score 1 · Answer 1 · answered Oct 31 '12 at 18:03

$A$ is onto, right? And $BA = 1$ implies $A$ is one-to-one, so $A$ is an automorphism. Multiplying by its inverse from the right gives $B = A^{-1}$, which implies $AB = 1$. No linearity was needed.

This is a simple statement of category theory: Every epi section is an iso and its retraction is its inverse. If $rs = 1$ and $\forall x, y :\ xs = ys \Rightarrow x=y$, then $srs = s$, so $sr = 1$ and $r$ and $s$ are inverse.

Proof on Endomorphism

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