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 $𝑃(N)=M$;
• $L(X,Y)$ means the set of all continuous linear operators from $X$ to $Y$.

Solution 1:

$(B)\Rightarrow (A)$. 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:

1. bounded, because both $P$ and $T$ are bounded;
2. injective, because $T$ is injective and the restriction of $P$ to $T(X)$ is the identity;
3. 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

1. $P$ is bounded, because $T$ and $S$ are bounded;
2. the range of $P$ is contained in $T(X)$;
3. 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.