Atkinson

If $T$ is Fredholm, then the operator $S$ of [A1] satisfies: $ST-I_X$ and $TS-I_Y$ have finite-dimensional ranges, hence are compact.

Conversely assume there is a bounded $S:Y\to X$ with $F:=ST-I_X$ and $G:=TS-I_Y$ compact. By boundedness choose a real $b\ge 0$ such that $\mathrm{\parallel }Sy\mathrm{\parallel }\le b\mathrm{\parallel }y\mathrm{\parallel }$ for every $y\in Y$, and put $C:=\max (b,1)>0$.

For every $x\in X$ one has $x=STx-Fx$, hence $\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }STx\mathrm{\parallel }+\mathrm{\parallel }Fx\mathrm{\parallel }\le b\mathrm{\parallel }Tx\mathrm{\parallel }+\mathrm{\parallel }Fx\mathrm{\parallel }\le C\mathrm{\parallel }Tx\mathrm{\parallel }+\mathrm{\parallel }Fx\mathrm{\parallel }$.

By [A3] the transpose of $G=TS-I_Y$ is $G^{\ast }=S^{\ast }{T}^{\ast }-I_{Y^{\ast }}$, so $S^{\ast }{T}^{\ast }=I_{Y^{\ast }}+G^{\ast }=I_{Y^{\ast }}-(-G^{\ast })$; for $g\in Y^{\ast }$ with $T^{\ast }g=0$ one has $S^{\ast }{T}^{\ast }g=S^{\ast }(T^{\ast }g)=0$ by linearity of $S^{\ast }$, hence $(I_{Y^{\ast }}-(-G^{\ast }))g=S^{\ast }{T}^{\ast }g=0$ and $\ker T^{\ast }\subseteq \ker (I_{Y^{\ast }}-(-G^{\ast }))$.

Under the hypothesis of [step 1.2], $\ker T$ is finite dimensional and $\mathrm{ran}T$ is closed, by [A2] applied to the estimate of [step 2.1] with the compact operator $F$.

Under the hypothesis of [step 1.2] the operator $G:=TS-I_Y$ is compact, so its transpose $G^{\ast }$ is compact by [A3]; the negative $-G^{\ast }$ is compact as well, because the image of a bounded set under $-G^{\ast }$ is the negative of its image under $G^{\ast }$ and negating a set preserves the compactness of its closure. So [A3] applies to the compact operator $-G^{\ast }$ and makes $\ker (I_{Y^{\ast }}-(-G^{\ast }))$ finite dimensional; by [step 2.2] the subspace $\ker T^{\ast }$ is finite dimensional.

Under the hypothesis of [step 1.2], the dual of the cokernel is finite dimensional: since $\mathrm{ran}T$ is closed by [step 3.1], [A4] gives $(\mathrm{coker}T{)}^{\ast }=(Y\mathrm{/}\mathrm{ran}T{)}^{\ast }\cong (\mathrm{ran}T{)}^{\perp }=\ker T^{\ast }$, which is finite dimensional by [step 3.2].

Under the hypothesis of [step 1.2], the cokernel is finite dimensional: if $Z$ is a normed space whose dual has ordered basis $h_1,\dots ,h_n$, then $\mathrm{\Psi }(z):=(h_1(z),\dots ,h_n(z))$ is linear and injective, because a nonzero $z$ has by [A5] a norm-one functional $h$, and $h={\sum }_i{c}_i{h}_i$ forces $\mathrm{\Psi }(z)\mathrm{\ne }0$; the inverse bijection carries an ordered basis of the finite-dimensional image $\mathrm{\Psi }(Z)$ to an ordered basis of $Z$ by [A5].

Under the hypothesis of [step 1.2] the operator $T$ is Fredholm, since its kernel is finite dimensional by [step 3.1], its range is closed by [step 3.1] and its cokernel is finite dimensional by [step 5.1]; with [step 1.1] this is the asserted equivalence.