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Let $V,V',V''$ and $W$ be vector spaces over $k$. Then, it is known that $\operatorname{Hom}(\cdot,V)$ is a contravariant exact functor, i.e. for each exact sequence

$0\to V'\to V\to V'' \to 0$,

and each $W$, the induced sequence

$0\to \operatorname{Hom}(V'',W)\to\operatorname{Hom}(V,W)\to\operatorname{Hom}(V',W)\to 0$

is exact.

But what if all spaces involved are Banach spaces (over $\mathbb{C}$ or $\mathbb{R}$) and if we replace $\operatorname{Hom}(\cdot,V)$ by $\operatorname{L}(\cdot,V)$, i.e. the continuous linear maps with domain $V$? Is this functor still exact? What if we restict ourselves even stronger to Hilbert spaces? I'm particularly interested in the specialization of $W=\mathbb{R}$ or $W=\mathbb{C}$, where we get the dual spaces (again, only continuous maps considered in the Banach- or Hilbert case).

I'm learning for an exam and the books I've been reading say nothing about it...

Thank you!

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In your first paragraph you write $\textrm{Hom}(V, \cdot)$ but describe $\textrm{Hom}(\cdot, W)$. There's a similar problem in your second paragraph. –  Zhen Lin Feb 11 '13 at 19:15
1  
Hint for $W = \mathbb{R}$ or $\mathbb{C}$: yes. Use Hahn-Banach. No for general $W$: math.stackexchange.com/q/96739 –  Martin Feb 11 '13 at 19:19
    
@Martin: This is not a comment, this is an answer. –  Martin Brandenburg Feb 11 '13 at 19:23
    
@Martin: Ok, will expand. –  Martin Feb 11 '13 at 19:27

2 Answers 2

up vote 3 down vote accepted

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

for proofs and further discussion of this last point.

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I decided to leave the Hilbert space case to you. The answer is yes and I can expand in case you are stuck. –  Martin Feb 11 '13 at 19:56
    
I guess there is a question as to how exactly you define the category of Banach spaces. It seems that you are considering the category whose morphisms are continuous linear maps with closed images, in which case the cokernel of a morphism is just the image of the map. But it is perhaps more natural to let the morphisms be all the continuous linear maps, in which case the cokernel of a morphism should be the closure of the image. This changes the answer for $L(\cdot, \mathbb{R})$, which I'll address in another answer. –  Nate Eldredge Feb 11 '13 at 22:11
    
@NateEldredge: no I'm using all the continuous linear maps. The morphisms with closed image do not form a category since they are not closed under composition. Why do you think that it seems that I use just those morphisms which are continuous linear maps with closed images? –  Martin Feb 11 '13 at 22:13
    
Notice that $0 \to V' \to V \to V'' \to 0$ exact means that $i$ is injective, $i(V) = \ker{p}$ and $p$ is surjective, so $i$ has automatically closed image (by exactness at $V$) and $p$ is (isomorphic to) the projection $V \to V/i(V')$, hence it is onto. –  Martin Feb 11 '13 at 22:21
    
I'm sorry, my category theory is really bad. You are right about the morphisms being all continuous linear maps. But what I mean is this: I would say that $0 \to V' \to V \to V'' \to 0$ being short exact should mean that $\operatorname{coker} i = \ker p$ and $\operatorname{coker} p = 0$. I believe the cokernel of a map $i$ should be, not its image (which need not be a Banach space if it is not closed), but the closure of its image. This means in order for the sequence to be exact, we only need that $\overline{(i(V))} = \ker p$ and $p$ has dense image. –  Nate Eldredge Feb 11 '13 at 22:26

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.

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