Is my proof showing $A$ and $B$ are invertible if $AB$ is invertible correct? $V$ is a finite dimensional vector space and $A, B \in \mathcal{L} (V)$ where $\mathcal{L}$ is the space of linear operators. The question precisely asks to prove the statement either ways (iff statement). I have no problem proving that if $A^{-1}$ and $B^{-1}$ exist $\implies$ $(AB)^{-1}$ exists and that $(AB)^{-1}=B^{-1}A^{-1}$. My question is for the backward direction, for which I have used to following argument:
If $(AB)^{-1}$ exists then $(AB)^{-1}AB=AB(AB)^{-1}=I$ 
Assume $S, T \in \mathcal{L}$  such that  $AS=SA=I$ and $BT=TB=I$, then $S=A^{-1}$ and $T=B^{-1}$ since the inverse is unique.
From the first equation we have $S=B(AB)^{-1}$ and $T=(AB)^{-1}A$, then
$A^{-1}$ and $B^{-1}$ exist, completing the proof in the backward direction.
Is my logic ok? Because I can't seem to find this answer anywhere.
 A: Hint: Consider, on the vector space of all sequences of real numbers, the transformation  $B$ that takes $u_1, u_2, u_3, \ldots$ to $0, u_1, u_2, u_3, \ldots$ (i.e., it prepends a zero) and $A$, which takes $u_1, u_2, u_3, \ldots$ to $u_2, u_3, \ldots$ (i.e., it strips off the first element). 
Then $AB = I$. But $A$ is not invertible. 
From this, you see that the hypothesis "$V$ is finite dimensional" is actually necessary. So perhaps start with asking yourself, "What main theorems do I know about transformations on finite dimensional vector spaces?" (... and about composite transformations, while you're at it.)
A: You have found a candidate for the inverse of $A$ to be $B(AB)^{-1}$.
This is, in fact, a right inverse since $A (B(AB)^{-1})=(AB)(AB)^{-1}=I$.
Now we need to use that the space is finite dimensional.
Since $A$ has a right inverse, its range is the whole space. Therefore its kernel is zero since the dimension of the space is finite.
This means that multiplication by $A$ is surjective and injective. Therefore it has an inverse.
For $B$ the argument is similar. $(AB)^{-1}A$ is a left inverse of $B$. Therefore the kernel of $B$ is zero. Using that the space is finite dimensional we get that the range is the whole space. Therefore it has an inverse too.
A: Note that this fails for infinite-dimensonal spaces $V$, e.g. take $V= \mathbb{R}^\mathbb{N}$ and define $A(x_0, x_1, x_2, \ldots) = (0, x_0, x_1, x_2, \ldots)$ and $B(x_0, x_1, x_2, x_3, \ldots) = (x_1 ,x_2, x_3, \ldots)$. Then $B \circ A$ is the identity but $A$ and $B$ are both not invertible, as $B$ is not onto and $A$ is not injective.
For finite dimensional spaces this canot happen as then $\dim(\operatorname{ker}(A)) + \dim(\operatorname{Im}(A)) = n$ for $A: V \to V$ linear and $\dim(V) = n$.
$A$ is injective iff $\operatorname{ker}(A) = \{0\}$ iff $\dim(\operatorname{Im}(A)) = n$ iff $A$ is onto.
And $AB$ invertible means $AB$ is injective and so $B$ is injective. (if $x \in \operatorname{ker}(B)$ then $x \in \operatorname{ker}(AB) = \{0\}$.) So $B$ is onto as well by the finite-dimensionality argument, hence invertible. So $A = (AB)B^{-1}$ is invertible as the product of invertible maps.
