l^-1∈L(w;v) is bijective?

Question

If l∈L(v;w) is bijective, then l^-1∈L(w;v) too? I'm not sure if it's this way that's why I ask, I'd appreciate your help.Until now I have done this... ”=⇒” Suppose T is invertible. To show that T is injective, suppose that u, v ∈ V are such that T (u) = T (v). Apply the inverse $$T^−1$$ of T to obtain $$ T^−1 T u = T^−1$$ T v so that u = v. Hence T is injective. To show that T is surjective, we need to show that for every w ∈ W there is a v ∈ V such that T v = w. Take $$v = T^−1w ∈ V$$ . Then $$T(T^−1w) = w$$. Hence T is surjective. ”⇐=” Suppose that T is injective and surjective. We need to show that T is invertible. We define a map S ∈ L(W, V ) as follows. Since T is surjective, we know that for every w ∈ W there exists a v ∈ V such that T v = w. Moreover, since T is injective, this v is uniquely determined. Hence define Sw = v. We claim that S is the inverse of T. Note that for all w ∈ W we have T Sw = T v = w so that T S = IW . Similarly for all v ∈ V we have ST v = Sw = v so that ST = IV . It remains to show that S is a linear map. For all w1, w2 ∈ W we have T(Sw1 + Sw2) = T Sw1 + T Sw2 = w1 + w2, so that Sw1 + Sw2 is the unique vector v in V such that T v = w1 + w2 = w. Hence Sw1 + Sw2 = v = Sw = S(w1 + w2).