Colimits in the category of (not necessarily locally convex) topological vector spaces

Question

Do colimits in the category of (not necessarily locally convex) topological vector spaces (over R, C, respectively) exist in general?

If no, is there a well-known condition of when they exist?

If yes, how can I describe the topology of the colimit?

(The example in my mind has following properties. First, it is a filtered colimit; it is a chain of inclusions, not necessarily linearly ordered, and not necessarily countable. Also, each inclusion is in general not a homeomorphism onto the image. Topology gets coarser and coarser as you embed into a larger space.)

(I'm also interested on a condition when a subset of the colimit is contained in one of the component spaces. For example, in the test function space $\mathcal{C}_c^\infty(\mathbb{R}^n)$ (which is of course NOT a colimit in the category of topological vector spaces), any bounded set is contained in one of $\mathcal{C}^{\infty}(K)$, $K$ a compact set. When can I say something similar?)

Answer 1

The Springer Lecture Notes 639 Topological Vector Spaces of Adasch, Ernst, and Keim contain in § 4 a more or less explicit construction of inductive (=co-) limits in the category of topological vector spaces based on the notion of a string: A sequence $(U_n)_{n\in\mathbb N}$ of balanced and absorbing sets such that $U_{n+1}+U_{n+1}\subseteq U_n$. Similarly to the locally convex theory (if $U$ is absolutely convex the sequence $U_n=2^{-n}U$ is a string) you can define a vector space topology by specifying a directed family of strings. If you have an inductive spectrum $(E_\alpha,i_\alpha)_{\alpha\in I}$ where $i_\alpha:E_\alpha\to E$ are linear mappings such the union of their ranges span $E$ you consider the family of all strings in $E$ all whose preimages are strings for the given topologies. The basics of such inductive limits are then rather similar to the locally convex case.

If the index set is countable and all $E_\alpha$ are locally convex the inductive limits in the categories of topological vector spaces and of locally convex spaces coincide -- but this seems to be the only case where you have "regularity results" on the behaviour of bounded or compact sets of the inductive limit like the one of Dieudonne-Schwartz or (if house advertising is permitted) in my Springer Lecture Notes Derived Functors in Functional Analysis.

Answer 2

I found a construction of the coproduct in the category of topological abelian groups from this reference: https://core.ac.uk/download/pdf/82771298.pdf, which can be also applied to the category of topological vector spaces without any difficulty.

The construction goes as follows.

1. First, form the algebraic direct sum $E:=\bigoplus_{i\in I}E_{i}$ of tvs's $E_{i}$'s. Consider the collection $\mathcal{P}$ of pairs $(N,\mathcal{T})$, where $N$ is a subspace of $E$ and $\mathcal{T}$ is a vector space topology on $E/N$ making the composition $E_{i}\rightarrow E\rightarrow(E/N,\mathcal{T})$ continuous.

2. Embed $E$ into the product space $\prod_{(N,\mathcal{T})\in\mathcal{P}}(E/N,\mathcal{T})$. Since the product space contains a copy of each $E_{i}$, this embedding is injective.

3. Then it is routine to check that $E$ endowed with the subspace topology is the coproduct.

As noted in comments, hence arbitrary colimits can be formed by passing to a quotient. However, the same construction perhaps can be still applied to arbitrary filtered colimits, though as noted in the reference this construction is only useful for showing existence and not really usable in practice.

Answer 3

In the note http://www-users.math.umn.edu/~garrett/m/fun/uncountable_coproducts.pdf it is proven that an uncountable (not-necessarily locally convex) coproduct of lines does not exist, by using the spaces $\ell^p$ with $0<p<1$, which are not locally convex.

EDIT/Correction: oops, yes, sorry, as in comments, what is proven there is not that no possible coproduct exists in the larger category, but that the locally convex coproduct is not a coproduct in the larger category.

Further, the device of using the non-locally-convex $\ell^p$ spaces does not seem to immediately give non-existence of a coproduct in the larger category (though I may be mistaken).

Answer 4

Indeed most texts concentrate on inductive limits LCS, since these are the most important in applications, but some results still holds for general TVS. Here are the basic ones, talking about sequences $(X_k,\mathcal U_k)$ of TVS (where $ \mathcal U_k$ denotes the set of nbd's of the origin of $X_k$), with continuous linear inclusion $X_k\subset X_{k+1}$. In particular, about the bornographical question.

• (Existence of the inductive limit). There is the strongest topology of TVS on $X_\infty:=\cup_{k\ge0}X_k$ that makes all $X_k\subset X_\infty$ continuous; a base of nbd's of $0$ is $\mathcal U_\infty:=\big\{\sum_{k\ge0}U_k: \forall k\ge0, U_k\in\mathcal U_k\big\}$. Here $\sum_{k\ge0}U_k:=\bigcup_{n\ge0}(U_1+\dots+U_n)$. For any TVS $Y$ and any linear map $L:X_\infty\to Y$, $L$ is continuous iff all restrictions $L_{|X_k}:X_k\to Y$ are continuous.

• Clearly, if all $X_k$ are LCS, so is $X_\infty$, meaning that for LCS the LCS inductive limit coincides with the TVS inductive limit.

• Dieudonné-Schwartz: for strict limits of TVS, i.e. $X_k\subset X_{k+1}$ is a subspace inclusion (i.e. $X_k$ has the induced topology from $X_{k+1}$), $X_k$ is then a subspace of $X_\infty$.

• Also, for strict limits: A subset $C\subset X_{n_0}$ is closed in $X_\infty$ iff it is closed in all $X_n$ for $n\ge n_0$. In particular (taking $C:=\{0\}$) if all $X_n$ are Hausdorff so is $X_\infty$ (which is not true for non-strict inductive limits even in LCS: check this spectacular example of the holomorphic germs by Jochen Wengenroth.

• Bounded sets: if each $X_k$ is a closed subspace of $X_{k+1}$, then any $A\subset X_\infty$ is bounded iff it is included in some $X_k$ and bounded there. This is clearly false without the closure assumption: take e.g. $X_k$ an increasing sequence of subspaces of $\ell_\infty$ starting from $X_0:=c_0$, all of them endowed with the $w^*$ topology of $\ell_\infty$. Then the unit ball $B$ of $X_0$ is bounded in $X_0$, hence in $X_\infty$ too, and so is its closure $\overline B^\infty$ in $X_\infty$. But $\overline B^\infty =\bigcup_{k\ge0}(\overline B^\infty\cap X_k)$ and $\overline B^\infty\cap X_k$ is the closure of $B$ in $X_k$, so it coincides with $\overline B^{w^*}\cap X_k$, and $\overline B^{\infty}=\overline B^{w^*}\cap X_\infty$; but by the Goldstine's theorem, $\overline B^{w^*}$ is the closed unit ball of $\ell_\infty$. Therefore $\overline B^{\infty}$ is bounded in $X_\infty$ but it is not included in any $X_k$ .

Answer 5

There is a general categorical notion of "topological categories", where for example the category of topological vector spaces is topological over the category of vector spaces (over $\mathbb{R}$ or $\mathbb{C}$). From the general theory of topological categories, cocompleteness of topological vector spaces can be deduced from cocompleteness of vector spaces. See for example Abstract and Concrete Categories (The Joy of Cats), Definition 21.1 and Theorem 21.16.

If you want to work instead with Hausdorff topological vector spaces, that too can be accommodated. See 25.2 example (5), 25.11, 25.15.