Under Dependent Choice, an injective bounded operator between Banach spaces has a bounded left inverse exactly when its range is closed and complemented

Assume $S$ is a bounded left inverse of $T$, so $ST=I_X$ by [L1]. If $T{x}_n\to y$ in $Y$, then $x_n=S(T{x}_n)\to S(y)$ because $S$ is bounded. Since $T$ is bounded as well, $T{x}_n\to T(Sy)$. Hence $y=T(Sy)$, so $\mathrm{ran}(T)$ is closed.

Conversely, assume $\mathrm{ran}(T)$ is closed and complemented in $Y$. Since $T$ is injective and has closed range, [L0] and [L3] make it bounded below. Thus the inverse $R:\mathrm{ran}(T)\to X$ defined by $R(Tx)=x$ is bounded.

With $P:=TS$, one has $P^2=T(ST)S=TS=P$. If $\mathrm{\parallel }Tx\mathrm{\parallel }\le C_T\mathrm{\parallel }x\mathrm{\parallel }$ and $\mathrm{\parallel }Sy\mathrm{\parallel }\le C_S\mathrm{\parallel }y\mathrm{\parallel }$, then $\mathrm{\parallel }Py\mathrm{\parallel }=\mathrm{\parallel }TSy\mathrm{\parallel }\le C_T{C}_S\mathrm{\parallel }y\mathrm{\parallel }$, so $P$ is bounded. For every $y\in Y$, $Py=TSy$ lies in $\mathrm{ran}(T)$. If $y=Tx$ is already in the range, then $Py=TSTx=Tx=y$. So $\mathrm{ran}(P)=\mathrm{ran}(T)$, and [L2] shows that the range is complemented.

Let $P:Y\to Y$ be a bounded projection onto $\mathrm{ran}(T)$, given by [L2]. Then $S:=R\circ P:Y\to X$ is bounded and $STx=R(P(Tx))=R(Tx)=x$ for every $x\in X$. Hence $S$ is a bounded left inverse of $T$.

Steps 2.1 and 2.2 prove the equivalence.