Is duality an exact functor on Banach spaces or Hilbert spaces?

To say that$\DeclareMathOperator{Hom}{Hom}$ $$ 0 \to V' \xrightarrow{i} V \xrightarrow{p} V'' \to 0 $$ is exact is equivalent to saying that $i$ is a kernel of $p$ and $p$ is a cokernel of $i$. This amounts to the automatic exactness of $$ 0 \to \Hom(W,V') \xrightarrow{i_\ast}\Hom(W,V) \xrightarrow{p_\ast} \Hom(W,V'') $$ and $$ 0 \to \Hom(V'',W) \xrightarrow{p^\ast} \Hom(V,W) \xrightarrow{i^\ast} \Hom(V',W) $$ for all Banach spaces $W$ where I write $\Hom(V,W) = L(V,W)$ for the space of continuous linear maps.

Consider the special case $W = \mathbb{R}$ (or $\mathbb{C}$):

Giving a morphism $f \colon \mathbb R \to X$ is the same as choosing a vector $x = f(1) \in X$ because for a scalar $\lambda$ we have $f(\lambda) = f(\lambda 1) = \lambda f(1) = \lambda x$. So: $\Hom(\mathbb R, X) = X$ for every Banach space and the sequence $$ 0 \to \Hom(\mathbb R,V') \xrightarrow{i_\ast}\Hom(\mathbb R,V) \xrightarrow{p_\ast} \Hom(\mathbb R,V'') \to 0 $$ really is the sequence $0 \to V' \xrightarrow{i} V \xrightarrow{p} V'' \to 0$ we started with.

Since $\operatorname{im} i = \ker{p}$, the map $i\colon V' \to V$ is a homeomorphism onto its range $i(V')$ and $i(V')$ is a closed subspace of $V$. A linear map $f' \colon V' \to \mathbb R$ thus corresponds to a linear functional on the subspace $i(V')$ of $V$ and the Hahn-Banach theorem allows us to extend that linear functional to all of $V$. In other words, $i^\ast \colon \Hom(V,\mathbb R) \to \Hom(V',\mathbb R)$ is always surjective and therefore the dual sequence $$ 0 \to (V'')^\ast \to V^\ast \to (V')^\ast \to 0 $$ is exact.

For general Banach spaces $W$, the answer is that neither $p_\ast$ nor $i^\ast$ need to be surjective: For $V' = c_0$ and $V = \ell_\infty$ and $V'' = \ell_\infty / c_0$, the sequence $$ 0 \to c_0 \xrightarrow{i} \ell_\infty \xrightarrow{p} \ell_\infty/c_0 \to 0 $$ is exact. Phillips' lemma says that for $W = c_0$ the identity $V' \to W$ cannot be extended to a morphism $V = \ell_\infty \to c_0$ and, equivalently, the identity $\ell_\infty/c_0 \to V''$ cannot be lifted to $\ell_\infty/c_0 \to \ell_\infty$.

See

• Do continuous linear functions between Banach spaces extend?
• Complement of $c_{0}$ in $\ell^{\infty}$

for proofs and further discussion of this last point.

I'm going to adjust your notation, because $V'$ looks too much like a dual space to me.

In the category of Banach spaces, where the morphisms are the continuous linear maps, one should perhaps interpret "image" in the categorical sense, as the closure of the image. (See the comments on Martin's answer.) If we interpret "exact" according to this sense, then a sequence $$0 \to X_1 \overset{S}{\to} X_2 \overset{T}{\to} X_3 \to 0$$ is exact iff $\ker S = 0$, $\overline{S(X_1)} = \ker T$, and $T(X_2)$ is dense in $X_3$.

In this setting, the functor $L(\cdot, W)$ is not exact, not even when $W = \mathbb{R}$. For instance, let's take $X_1 = \ell^1$, $X_2 = \ell^2$, $S$ the inclusion map, and $X_3 = 0$. Since $S$ is injective and has dense image, this sequence is exact. If $L(\cdot, \mathbb{R})$ were an exact functor, then $S^* : (\ell^2)^* \to (\ell_1)^*$ should also be injective and have dense image. $S^*$ is injective, but since it is a map from the separable Banach space $(\ell^2)^* = \ell^2$ to the non-separable Banach space $(\ell^1)^* = \ell^\infty$, it cannot have dense image. (In fact, $S^*$ is just the inclusion map $\ell^2 \to \ell^\infty$ and the closure of $\ell^2$ in $\ell^\infty $ is $c_0$, the sequences which vanish at infinity.)

If you work in the category of reflexive Banach spaces, then $L(\cdot, \mathbb{R})$ is an exact functor; this is a fairly straightforward diagram chase using the Hahn-Banach theorem.