# Complex of banach spaces is exact if and only if its dual is exact

Let's consider two complexes of Banach spaces: $X \rightarrow Y \rightarrow Z$, with the maps $S: X \rightarrow Y$, $T: Y \rightarrow Z$. The dual complex looks like $Z^{*} \rightarrow Y^{* } \rightarrow X^{*}$, and the maps are their Banach adjoints. Also, it's given that $TS=0$.

I would like to proof the following statement: If $im(S)$ and $im(T)$ are closed, then $(ker(T)/im(S))^{*} = ker(S^{*})/im(T^{*})$. ( this is all about the duality between homology and cohomology in good cases).

I suppose that following lemma may help to reach the final result: complex of banach spaces is exact if and only if its dual is exact (this is done under some assumption, exactly, that $im(T)$ is closed (this should imply that $im(S^{*})$ is closed.)

The questions are: 1) What is the possible approach to prove the lemma (looks as if it can be proven using such kind of isometric isomorphisms as $(coker(T^{*}))=(ker(T))^{*}$, $ker(T^{*}) = (coker(T))^{*}$, but after considerable amount of time spent on it, i'm not sure in the final success).

2) Does the lemma really help to derive the main statement? Looks as if it is impossible to proceed without some stuff from homological algebra, but it would rather entertaining to find out the exact solution without some extra machinery (so far, playing with isometric isomorphisms didn't help much, i go circles by circles all around the same point).

Any sort of help would be much appreciated.

Consider the complex and its dual: $$X \overset{S}{\rightarrow} Y \overset{T}{\rightarrow} Z\\ X^*\overset{S^*}{\leftarrow} Y^*\overset{T^*}{\leftarrow} Z^*$$ We have $S^* \circ T^* = (T\circ S)^* = 0^* = 0$ so indeed the second one is a complex. Let's show exactness: consider $\phi\in Y^*$ so that $S^*(\phi) = 0$, that is $\phi(S X) = 0$. Now, $S X = \textrm{Ker} T$. Therefore, $\phi$ is $0$ on $\textrm{Ker} T$. Let us now consider the maps $T \colon Y \to X$ and $\phi \colon Y \to k$. The kernel of $\phi$ contains the kernel of $T$. Therefore, there exists a (unique) linear map $\psi$ from the image of $T$ to $k$ so that $\psi \circ T = \phi$. This map $\psi$ will be continuous in fact. The reason is that by the open mapping theorem for Banach spaces we have $Y/\ker T \simeq Im T$ ( we are using the fact that $Y$, $Z$ are Banach and $Im T$ is closed in $Z$, so also Banach). Now, since $k$ are the numerics ( $\mathbb{R}$ or $\mathbb{C}$) we can extend $\psi$ to a continuous functional on the whole $Z$ ( by Banach-Steinhaus).