I should say that I'm not a category theorist or an abstract algebraist, so maybe this will be very pedestrian. I have the following, somewhat vague question:

I have categories C and D, a forgetful functor $U:C\rightarrow D$. This has a left adjoint, but does not have a right adjoint. Are there other situations where this occurs, and what "workarounds" are there.

Let's be more precise with an example which is close to my actual situation. Let $A$ be a $k$-algebra ($k=\mathbb C$ if you like) and let $U:A{\sf -mod} \rightarrow k{\sf -vect}$ the forgetful functor from left $A$-modules to $k$-vector spaces. This has a left adjoint $A \otimes \underline{\ \ }$; you have a natural bijection, for a vector space $V$ and a module $M$,

\[ \text{Hom}_{k{\sf -vect}}(V,U(M)) \cong \text{Hom}_{A{\sf-mod}}(A\otimes V,M), \]

which sends $T:V\rightarrow U(M)$ to $a\otimes x\mapsto a\cdot T(x)$. Then a module of the form $A\otimes V$ is "free", and this leads to consideration of projective modules etc. In my situation (dealing with topological algebras etc.) I can parallel all of this.

Similarly, we have a right adjoint $\text{Hom}_{k{\sf -vect}}(A,\underline{\ \ })$. This gives a natural bijection

$$ \text{Hom}_{A{\sf -mod}}(M,\text{Hom}_{k{\sf -vect}}(A,V)) \cong \text{Hom}_{k{\sf -vect}}(U(M),V), $$ which identifies $T:U(M)\rightarrow V$ with $x \mapsto (a\mapsto T(a\cdot x))$. Then one might call a module of the form $\text{Hom}_{k{\sf -vect}}(A,V)$ "cofree", and this leads to injective modules, etc.

In my situation, my category of modules is not a "closed category"; if cofree modules existed, the naturality of the above bijection would force them to $\text{Hom}(A,V)$ (for a suitable meaning of $\text{Hom}$), but this object is not in my category. I'm hence missing a good notion of "cofree"; but really I'm interested in injectives. Are there situations (say, in algebra) where something similar occurs?

Edit: People demand a more explicit example. Suppose $A$ is a von Neumann algebra, and my category is those modules which are dual Banach spaces $E$, with a bounded left action of $A$, and such that for each $x\in E$, the orbit map $A\rightarrow E; x\mapsto a\cdot x$ is weak$^*$-continuous. The forgetful functor is to the category of dual Banach spaces with weak$^*$-continuous bounded linear maps as morphisms. The right adjoint "should be" $B^\sigma(A,\underline{\ \ })$, the weak$^*$-cts bounded linear maps. But this is rarely itself even a dual Banach space.

  • $\begingroup$ I don't understand the tags "functional analysis" and "cohomology". Also I don't see a connection between the question and the example. $\endgroup$ Mar 2, 2011 at 11:41
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    $\begingroup$ Sorry for the downvote, but I actually think you should be much more precise about this. Does this happen? Yes, all the time! The forgetful functor from groups to sets has a left adjoint but not a right adjoint, and this is extremely common. Is there a workaround? I don't know: workaround for what exactly? Instead of the highly elliptical "dealing with topological algebras, etc.", please tell us exactly what you are dealing with, and then maybe one can tell if there are injectives or whatever, and maybe you can tell us what you want to do if there aren't. $\endgroup$
    – Todd Trimble
    Mar 2, 2011 at 11:57
  • $\begingroup$ Okay, I've edited. But I really, really would just like to learn about similar situations, rather than get very specific answers. If that's not appropriate for MO, then I'll delete the question I guess... $\endgroup$ Mar 2, 2011 at 12:24
  • $\begingroup$ @Matthew: the problem as I see it is that the displayed question is at a level of extreme generality, and it's very hard to tell what would be a helpful reply. Your edit helps a bit; my intent was to try to reach a level of more useful generality. I will make some sort of stab at reply, with the caveat that I know a lot more about category theory than I do about functional analysis. $\endgroup$
    – Todd Trimble
    Mar 2, 2011 at 13:54
  • $\begingroup$ Or maybe I'll hold off on a reply since Theo has entered the room. I'm curious how you will reply (Matthew). $\endgroup$
    – Todd Trimble
    Mar 2, 2011 at 13:55

2 Answers 2


I did say I would respond, so here are some thoughts. I'll repeat the caveat that I am not a functional analyst by any means.

There's something in category theory called the Chu construction, which has very nice categorical properties and which seems to be a natural place in which to embed your category of dual Banach spaces and weak-* continuous maps, and also an environment in which cofree module constructions could be carried out.

Let $\mathbf{B}$ be a symmetric monoidal closed complete and cocomplete category, (in our situation, $\mathbf{B}$ will be the category of Banach spaces) and let $D$ be a putative "dualizing object" (which here will be the scalar field, let's say $\mathbb{C}$ for complex Banach spaces). The Chu construction $Chu(\mathbf{B}, D)$ is the category whose objects are triples $(B, C, \phi: B \otimes C \to D)$ and whose morphisms $(B, C, \phi) \to (B', C', \phi')$ consist of pairs of maps $(f: B \to B', g: C' \to C)$ such that

$$\phi'\circ (f \otimes 1_{C'}) = \phi \circ (1_B \otimes g)$$

There is a certain canonical symmetric monoidal closed structure on $Chu(\mathbf{B}, D)$ such that every object of $Chu$ is reflexive with respect to the triple $(I, D, \lambda: I \otimes D \to D)$, where $I$ is the monoidal unit of $\mathbf{B}$ and $\lambda$ is the canonical action of the unit. (The dual of an object $(B, C, \phi: B \otimes C \to D)$, by homming into $(I, D, \lambda)$, is just $(C, B, \phi \circ \sigma)$ where $\sigma: C \otimes B \to B \otimes C$ is the canonical symmetry.) The Chu category is also complete and cocomplete, if $\mathbf{B}$ is.

There is also an embedding $i: \mathbf{B} \to Chu(\mathbf{B}, D)$ taking an object $B$ to the triple $(B, \hom(B, D), eval: B \otimes \hom(B, D) \to D)$. This embedding preserves the tensor product up to canonical isomorphism.

Now let $C'$, $C$ be dual Banach spaces: $C = \hom(B, D)$ as above. Theo quotes in one of his comments the fact that a map $g: C' \to C$ between dual Banach spaces is an adjoint, $g = f^\ast$ for some $f: B \to B'$, if and only if $g$ is weak-$\ast$ continuous. (I hope I've got that right.) So if $DualBan$ is the category of dual Banach spaces and weak-$\ast$ continuous maps, we have another embedding

$$j: DualBan \to Chu(Ban, \mathbb{C})$$

which takes a dual Banach space $C$ to the triple $(C, B, eval: C \otimes B \to \mathbb{C})$ where $B$ is the predual of $C$; it takes a weak-$\ast$ continuous map $f$ to the pair $(f, g)$, where $g$ is the adjoint to $f$. The embedding $j$ is full and faithful.

The point I'm driving at is that if your category $DualBan$ lacks certain good properties (for purposes of co-free modules, etc.), you can nevertheless fully embed it in the very nice category $Chu(Ban, \mathbb{C})$ where the co-free modules are available.

For example, if $A$ is a Von Neumann algebra with predual $A^\flat$, there is a monoid object $j(A) = (A, A^\flat, eval: A \otimes A^\flat \to \mathbb{C})$ in $Chu = Chu(Ban, \mathbb{C})$. One could study the category of Chu-modules over $j(A)$, i.e., the category of monoid actions of $j(A)$ in the monoidal category $Chu$.

Since $Chu$ is symmetric monoidal closed, one can mimic the usual constructions from commutative algebra, and in particular form the internal hom $\hom_{Chu}(A, M)$ for any Chu-module $M$ over $A$. So the forgetful functor

$$ChuMod_A \to Chu$$

would have both left and right adjoints.

In summary, this is following a philosophy of Grothendieck: instead of working with a category of 'nice' objects but with bad categorical properties, try expanding to a wider context where the objects might be 'wilder' (e.g., Chu triples) but which has good categorical properties. It is certainly my categorical reflex in this case.

Edit: Here are some references. The Chu category I described is called the category of coherent Banach spaces by Girard, who applies them to the semantics of his linear logic. The Chu construction is very well-known in category theory, and has been studied extensively by Michael Barr and Vaughan Pratt (just to name two authors; Barr was Chu's adviser for his master's thesis, where this construction first appeared). In answer to Yemon, Barr does mention the connection with Mackey spaces in numerous places, for example here, and the references therein.

The basic idea is that in functional analysis, there is a multiplicity of possible topologies (weak, weak-$\ast$, Mackey, etc.), and the Chu construction provides an environment for discussing all of them within a single category, much as Theo surmised in one of his comments below. As I understand it, similar considerations were paramount in Barr's mind when he was doing seminal work on $\ast$-autonomous categories (see his book, Springer Lecture Notes in Mathematics 752).

  • $\begingroup$ Thank you very much! You've got the correct part of my comment right and, as far as I can tell from a first cursory reading, all the things you're saying seem to make sense from the functional analytic perspective (and some things ring vaguely familiar). For me it seems more familiar at the moment to think of a Chu-Banach space as a pair of Banach spaces together with a bilinear map $\langle \cdot, \cdot, \rangle: B \times C \to \mathbb{C}$ and the embedding $\mathbf{B} \to Chu(\mathbf{B})$ is simply sending $B$ to the usual duality paring $B,B^{\ast}$. One caveat: Matthew seems to work in the $\endgroup$ Mar 2, 2011 at 22:34
  • $\begingroup$ internal category of $\mathbf{B}$. More explicitly: with the category of Banach spaces and bounded linear maps (which is additive but only finitely complete and cocomplete). $\endgroup$ Mar 2, 2011 at 22:38
  • $\begingroup$ Thanks! And thanks for that caveat; it's of course true that the category of Banach spaces and maps of norm bounded above by 1 is a complete and cocomplete category, and probably I was unconsciously using that instead. $\endgroup$
    – Todd Trimble
    Mar 2, 2011 at 22:47
  • $\begingroup$ Theo (and Todd): do you know if this is related to the Mackey topology on TVSes? en.wikipedia.org/wiki/Mackey_topology I have a recollection of seeing the phrase or something similar come up in work of Michael Barr, who I know has done things with relatives of the Chu construction $\endgroup$
    – Yemon Choi
    Mar 2, 2011 at 23:02
  • $\begingroup$ I don't see the Mackey topology entering here (for Banach spaces it is the usual norm topology). But googling for the obvious yields Barr's paper on $\ast$-autonomous categories and topological vector spaces, so there's certainly a relation ftp.math.mcgill.ca/pub/barr/pdffiles/tvs.pdf From the nlab I learned that Chu is a student of Barr. $\endgroup$ Mar 2, 2011 at 23:16

This is a long comment (so an answer) which is really just a thank you to Todd Trimble-- this idea of Chu duality (or perhaps, Barr-Chu duality) seems really interesting. For Banach spaces, it basically captures the idea of a "dual pair", which is certainly an idea out there in the literature. But the key difference seems (as ever with category theory!) to worry about the morphisms.

So an object is $(E_1,E_2,\langle\cdot,\cdot,\rangle)$ which is a pair of Banach spaces together a bilinear pairing to $\mathbb C$. A morphism $(E_1,E_2,\langle\cdot,\cdot,\rangle)\rightarrow(F_1,F_2,\langle\cdot,\cdot,\rangle)$ is a pair of maps $f:E_1\rightarrow F_1$ and $g:F_2\rightarrow E_2$ which commutes across the pairing: $$ \langle f(x), \mu \rangle = \langle x, g(\mu) \rangle \qquad (x\in E_1, \mu\in F_2). $$

Then you can carry out tensor product, and internal hom-space, constructions. What's remarkable, to me, is that the obvious notion of a module in this setting automatically gives me this concept of "normality"-- something which has always seemed convenient, but slightly ad-hoc. So for that alone, this is a nice discovery!

  • $\begingroup$ Thanks, Matthew -- it's my pleasure, of course. Category theorists and logicians have spent a lot of time with the Chu construction, but I am not aware of practicing analysts doing so, and I think it would be great to see what they can do with the idea. Good luck! $\endgroup$
    – Todd Trimble
    Mar 8, 2011 at 2:38

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