It's well-known that the category $\mathsf{Ban}_c$ of Banach spaces and linear contractions (i.e. of norm $\leq 1$) is $\omega_1$-accessible. It is also (co)complete, and hence locally $\omega_1$-presentable. This is typically proven in any reference book on accessible categories.

The larger category $\mathsf{Ban}$ of Banach spaces and all bounded linear maps lacks all infinite coproducts (unless all but finitely many factors are 0 -- see here for a proof by Martin Brandenburg that not all coproducts exist, reflecting on the proof yields this conclusion). So $\mathsf{Ban}$ is not locally presentable. But it might still be $\omega_1$-accessible.

One piece of evidence for accessibility is that $\ell_1(\omega)$ is a dense generator in $\mathsf{Ban}$ just as it is in $\mathsf{Ban}_c$: $\ell_1(\omega)$ corepresents sequences $(x_i)_i$ such that $\sum \|x_1\|$ converges, and it has split subobjects corepresenting addition and scalar multiplication, so any natural transformation $\mathsf{Ban}(i_{\ell_1(\omega)}-,X) \implies \mathsf{Ban}(i_{\ell_1(\omega)}-,Y)$ [where $i_{\ell_1(\omega)}$ is the inclusion functor for the full subcategory on $\ell_1(\omega)$] must come from a linear map preserving absolutely convergent sequences, which is true if and only if the map is bounded.

But I haven't been able to decide whether or not $\mathsf{Ban}$ has $\omega_1$-filtered colimits. I believe I have convinced myself that the inclusion $\mathsf{Ban}_c \to \mathsf{Ban}$ preserves $\omega_1$-filtered colimits, because any cocone on an $\omega_1$-filtered diagram of contractions has bounded norms on its legs, and so we can divide through by a constant to obtain an "equivalent" cocone of contractions. It would be nice to replace an $\omega_1$-filtered diagram in $\mathsf{Ban}$ with one in $\mathsf{Ban}_c$, but I'm not sure if this is possible in general.

If, optimistically, $\omega_1$-filtered colimits do exist, the next step is to determine the $\omega_1$-presentable objects and whether everything is an $\omega_1$-filtered colimit thereof, but this is even more obscure to me.

  • 1
    $\begingroup$ Re your final paragraph, the obvious guess is that $\omega_1$-presentable objects are separable Banach spaces and every Banach space is the colimit of its separable subspaces. $\endgroup$ – Eric Wofsey Sep 7 '15 at 5:04
  • $\begingroup$ Okay -- the closed separable subspaces of a Banach space form an $\omega_1$-filtered diagram in $\mathsf{Ban}_c$ whose colimit can be computed as in $\mathsf{Set}$ to be the original space. And I suppose the presentability question could at least be explored in $\mathsf{Ban}_c$ to start with. $\endgroup$ – tcamps Sep 7 '15 at 7:04

Okay, the answer is yes, $\mathsf{Ban}$ is $\omega_1$-accessible. Below I argue that $\mathsf{Ban}$ has $\omega_1$-filtered colimits, computed by modifying diagrams to lie in $\mathsf{Ban}_c$ and using that $\mathsf{Ban}_c \to \mathsf{Ban}$ preserves $\omega_1$-filtered colimits. The replacement is such that any functor $\mathsf{Ban} \to \mathcal{C}$ preserves $\omega_1$-filtered colimits if and only if its restriction to $\mathsf{Ban}_c$ does. In particular, the $\omega_1$-presentable objects of $\mathsf{Ban}$ coincide with those of $\mathsf{Ban}_c$. Every object in $\mathsf{Ban}$ is an $\omega_1$-filtered colimit of these because it is so in $\mathsf{Ban}_c$ and the colimits are preserved. Hence $\mathsf{Ban}$ is $\omega_1$-accessible, and $\mathsf{Ban}_c \to \mathsf{Ban}$ is a (non-full) $\omega_1$-accessible embedding.

This begs the question: what is a good sketch or axiomatization for $\mathsf{Ban}$ as a category of models and homomorphisms? (Okay, using $\mathsf{Ban}(\ell_1(\omega),-)$ to give an underlying set, we can presumably do this by axiomatizing a set equipped with a set of countable subsets to be interpreted as the sets $(x_i)_{i \in \mathbb{N}}$ such that "$\sum_i \|x_i\|$ is finite").

Also, this method does not explicitly identify the $\omega_1$-presentable objects in $\mathsf{Ban}$ (equivalently, in $\mathsf{Ban}_c$). Since $\ell_1(\omega)$ is a $\omega_1$-presentable strong generator, they are given by countable colimits of $\ell_1(\omega)$. This probably results in exactly the separable Banach spaces as Eric Wofsey suggests, but I haven't checked.

For completeness, the argument that $\mathsf{Ban}_c \to \mathsf{Ban}$ preserves $\omega_1$-filtered colimits goes as follows. If $X: I \to \mathsf{Ban}_c$ is an $\omega_1$-filtered diagram, and $f_i: X_i \to Y$ is a cocone in $\mathsf{Ban}$, then I claim that the norms of the legs of $f$ are bounded. Otherwise, we could choose $(i_n)_{n \in \mathbb{N}}$ and $x_n \in X_{i_n}$ with $\|x_n\| \leq 2^{-n}$ but $\|f_{i_n}(x_n)\| \geq 3^n$. By $\omega_1$-filteredness, we can find an $\alpha \in I$ with $\alpha\geq i_n$ for each $n$. Then $\sum X_{i_n\alpha}(x_n)$ converges absolutely, but under $f_\alpha$ it is sent to $\sum_n f_{i_n}(x_n)$ which must diverge, contradicting the boundedness of $f_\alpha$.

So there is $B$ such that $\|f_i\| \leq B$ for all $i \in I$. Then $f/B$ is a cocone in $\mathsf{Ban}_c$. So, letting $(\varinjlim X,(\eta_i)_{i \in I})$ denote the colimit cocone in $\mathsf{Ban}_c$, there is a unique cocone map $\bar f: (\varinjlim X (\eta_i)_{i \in I}) \to (Y,f/B)$ in $\mathsf{Ban}_c$, and $Bf$ is the unique cocone map $(\varinjlim X (\eta_i)_{i \in I}) \to (Y,f)$ in $\mathsf{Ban}$, and $\varinjlim X$ is the colimit

Here's how to reduce $\omega_1$-filtered colimits in $\mathsf{Ban}$ to those in $\mathsf{Ban}_c$. It suffices to treat $\omega_1$-directed colimits (since every $\lambda$-filtered diagram has a cofinal $\lambda$-directed subdiagram). If $X: I \to \mathsf{Ban}$ is such a diagram, first observe that $K_i := \{x \in X_i \mid \exists j \geq i \, X_{ij}(x) = 0\}$ forms a closed subspace of $X_i$ (this uses $\omega_1$-filteredness of $I$). So we can let $X_i' = X_i/K_i$, and then $X$ descends to a functor $X'$ in the obvious way, with an isomorphic category of cocones. Now $X'$ has the property that

(*) For any $x \in X_i'$ and $j \geq i$, $X_{ij}'(x) = 0$ iff $x=0$.

Pick $X_{i_0} \neq 0$ (otherwise the colimit is just 0), let $I_{\geq i_0} = \{i \in I \mid i \geq i_0\}$, which is cofinal by filteredness of $I$. Then let $X'': I_{\geq i_0} \to \mathsf{Ban}$ be the restriction of $X'$ to $I_{\geq i_0}$. Now we let $X'''_i=X''_i$ but $X'''_{ij} = \frac{\|X''_{i_0i}\|}{\|X''_{i_0j}\|}X''_{ij}$, using (*) to ensure that the denominators are nonzero (this is the only step that requires the indexing category to be a poset, we could have waited to pass to a cofinal poset a this point). Then $X'''\cong X''$, but lies in $\mathsf{Ban}_c$; we can take its colimit there; the colimit is preserved by the inclusion $\mathsf{Ban}_c \to \mathsf{Ban}$, and $\varinjlim X''' = \varinjlim X'' = \varinjlim X' = \varinjlim X$ so we have our desired colimit.

Oh, and here's an argument that $\ell_1(\omega)$ is $\omega_1$-presentable. $\omega_1$-filtered colimits in $\mathsf{Ban}_c$ are computed as in $\mathsf{Vect}$, the norm being maximal such that structure maps are contractions. As discussed in the question, $\ell_1(\omega)$ corepresents sequences whose norms sum to $\leq 1$. If $(x_n)_{n \in \mathbb{N}}$ is a sequence in $\in \varinjlim X$ with $\sum_n \|x_n\|_{\varinjlim X} = a \leq 1$, then by $\omega_1$-filteredness, we can find an $X_i$ with all the $x_n \in X_i$. Moreover, we can find a sequence $(X_{i_k})_{k \in \mathbb{N}}$ with $\sum_n \|x_n\|_{X_{i_k}} \to a$ as $k \to \infty$. Then by $\omega_1$-filteredness again, we can find $X_j$ with $j\geq i_k$ for all $k$, and it follows that $\sum_n \|x_n\|_{X_j} = a \leq 1$. So the map $\varinjlim \mathsf{Ban}_c(\ell_1(\omega),X) \to \mathsf{Ban}_c(\ell_1(\omega),\varinjlim X)$ is surjective. It is also injective: If $(x_n)_{n \in \mathbb{N}} = (0)_{n \in \mathbb{N}}$ in $\varinjlim X$, then each $x_n = 0$ already at some stage $i_n$, so by $\omega_1$-filteredness, each $x_n = 0$ in some $X_j$ with $j\geq i_n$ for all $n$. So $x_n = 0$ in $\varinjlim \mathsf{Ban}_c(\ell_1(\omega),X)$


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