Under Dependent Choice, a surjective bounded operator between Banach spaces has a bounded right inverse exactly when its kernel is complemented

Assume $S$ is a bounded right inverse of $T$, so $TS=I_Y$ by [L1]. Define $P:=I_X-ST$. Then $P^2=I_X-2ST+STST=I_X-ST=P$, because $TST=T$. Also, if $\mathrm{\parallel }Sy\mathrm{\parallel }\le C_S\mathrm{\parallel }y\mathrm{\parallel }$ and $\mathrm{\parallel }Tx\mathrm{\parallel }\le C_T\mathrm{\parallel }x\mathrm{\parallel }$, then $\mathrm{\parallel }STx\mathrm{\parallel }\le C_S{C}_T\mathrm{\parallel }x\mathrm{\parallel }$ and therefore $\mathrm{\parallel }Px\mathrm{\parallel }\le (1+C_S{C}_T)\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in X$; so $P$ is a bounded projection. Finally, $TP=T-TST=0$, so $\mathrm{ran}(P)\subseteq \ker T$.

Conversely, assume $\ker T$ is complemented. Then there is a closed subspace $L\le X$ with $X=\ker T\oplus L$. The restriction $T{\mathrm{\mid }}_L:L\to Y$ is injective, because $L\cap \ker T=\{0\}$, and it is surjective because every $x\in X$ decomposes as $k+\mathrm{\ell }$ with $Tx=T\mathrm{\ell }$. Since $L$ is closed in the Banach space $X$, [L4] makes $L$ Banach.

If $x\in \ker T$, then $Px=x-STx=x$. Hence $\ker T\subseteq \mathrm{ran}(P)$, and step 1.1 gives $\mathrm{ran}(P)=\ker T$. Therefore [L2] makes $\ker T$ complemented.

The map $T{\mathrm{\mid }}_L$ is a bounded bijection from the Banach space $L$ onto the Banach space $Y$, so [L0] and [L3] make it bounded below. Hence its inverse $R:Y\to L$ is bounded, because $\mathrm{\parallel }R(y)\mathrm{\parallel }\le c^{-1}\mathrm{\parallel }y\mathrm{\parallel }$ when $\mathrm{\parallel }T\mathrm{\ell }\mathrm{\parallel }\ge c\mathrm{\parallel }\mathrm{\ell }\mathrm{\parallel }$. The inclusion $L\hookrightarrow X$ now gives a bounded linear map $S:Y\to X$ with $TS=I_Y$. Thus $S$ is a bounded right inverse.

Steps 2.1 and 2.2 prove the equivalence.