Given $T \in L(X,Y)$, show the equivalence between: existence of $S$ such that $S(T(x))=x$, and $T$ being injective with $T(X)$ complemented in $Y$ Given $X,Y$ Banach spaces and $T \in L(X,Y)$, show that the following sentences are equivalent:
A) there exists $S \in L(Y,X)$ such that $S(T(x))=x$ for all $x \in X$.
B) $T$ is injective and $T(X)$ is a complemented space of $Y$.

Context:
I was given this exercise in my functional analysis course but I don’t know how to solve this.
All I have understood so far in this exercise are the following:


*

*I was given the following definition of "complemented space'': a closed subspace $M$ is complemented in $N$ if exists a topological complement of $M$ in $N$ or equivalently if there exists a linear continuous projection $P$ in $N$ such that $(N)=M$;

*$L(X,Y)$ means the set of all continuous linear operators from $X$ to $Y$.

 A: $(B)\Rightarrow (A)$. Since $T(X)$ is topologically complemented,  let $P:Y\to T(X)$ be a bounded linear projection. Then the operator 
$$P \circ T: X\to T(X)$$
is:


*

*bounded, because both $P$ and $T$ are bounded;

*injective, because $T$ is injective and the restriction of $P$ to $T(X)$ is the identity;

*surjective, again because the restriction of $P$ to $T(X)$ is the identity.


Moreover, $T(X)$ is closed, so it is a Banach space. Hence by the inverse mapping theorem, $P\circ T$ has a bounded linear inverse $(P\circ T)^{-1}:T(X)\to X$. We now define an operator $S:Y\to X$ by
$$S:=(P\circ T)^{-1}P $$
then $S$ is bounded, and $S\circ T=(P\circ T)^{-1}P\circ T=I$ so that $STx=x$ for all $x\in X$ as required.
$(A)\Rightarrow (B)$. Suppose that $Tx=Ty$. Then $x=STx=STy=y$. So $T$ is injective.
Since $S$ is bounded, we have
$$\|x\|=\|STx\|\leq \|S\|\|Tx\| $$
Thus 
$$\|Tx\|\geq \|S\|^{-1}\|x\|\qquad \forall x\in X $$
i.e. $T$ is also bounded from below (other than from above), and hence it is an isomorphism between $X$ and its range $T(X)$. In particular $T(X)$ is a Banach space and therefore it is closed.
It remains to find a continuous linear projection from $Y$ to $T(X)$. Define
$$P:=TS $$
Then 


*

*$P$ is bounded, because $T$ and $S$ are bounded;

*the range of $P$ is contained in $T(X)$;

*for all $y=Tx\in T(X)$, we have 
$$Py=P(Tx)=T(STx)=Tx $$
so that the restriction of $P$ to $T(X)$ is the identity. This completes the proof.

