invertible if and only if bijective

Solution 1:

From a category perspective, the natural definition for $T$ to be invertible here is that there exists $S\in L(W,W)$ such that $S\circ T=T\circ S=I$.

As usual, $S\circ T=I$ implies that $T$ is injective, and $T\circ S=I$ entails that $T$ is surjective. So $T$ invertible implies it is bijective.

Now assume that $T$ is bijective. Then we have $S=T^{-1}$ that fits the definition above, provided we can prove that it belongs to $L(W,W)$.

Let us write that $T$ is linear, namely $$T(\lambda x+\mu y)=\lambda T(x)+\mu T(y)$$ for all $x,y\in W$, and $\lambda,\mu\in\mathbb{R}$. This is true in particular for $x=T^{-1}(x')$ and $y=T^{-1}(y')$ for all $x',y'\in W$. So $$ T(\lambda T^{-1}(x')+\mu T^{-1}(y'))=\lambda TT^{-1}(x')+\mu TT^{-1}(y')=\lambda x'+\mu y'. $$ It only remains to apply $T^{-1}$ to the LHS and the RHS of the above. This proves the linearity of $T^{-1}$.

So yes, bijectivity is equivalent to invertibility in $L(W,W)$.