The injections (coprojections) of a coproduct in a category are very often monomorphisms. For instance, this happens in any extensive category (essentially by definition) and also in any category with zero morphisms (since in that case they are split monos). However, there are examples of (complete and cocomplete) categories where this is not always the case, e.g. in the category of commutative rings the coproduct is the tensor product, and the injection $\mathbb{Z} \to \mathbb{Z} \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2$ is not monic.

My question is, can you give an example of a complete and cocomplete closed monoidal category (a "Benabou cosmos") in which coproduct injections are not always monic? (Or, I suppose, a proof that no such example exists, but that would surprise me.)

  • $\begingroup$ Do you know that commutative rings (or, for that matter, $Set^{op}$) do not admit any closed monoidal structure? $\endgroup$ – Eric Wofsey Jun 3 '15 at 22:21
  • $\begingroup$ @EricWofsey No, I don't. (You're right, $\mathrm{Set}^{\mathrm{op}}$ is also a fine example, e.g. $\emptyset \times X \to X$ is not epic if $X$ is nonempty.) So if you could give a closed monoidal structure on either of them, that would answer the question. $\endgroup$ – Mike Shulman Jun 3 '15 at 22:44
  • 1
    $\begingroup$ For what it's worth, $\mathrm{Set}^{\mathrm{op}}$ has no closed monoidal structure. Just by playing with the adjunction you can see that the unit of such a structure has to be a finite set and, if it has $m$ elements, then the number of functions $X \to Y$ would have to be a power of $m$ for any two finite sets $X$ and $Y$. $\endgroup$ – Karol Szumiło Jun 3 '15 at 23:40
  • 2
    $\begingroup$ More generally, here's something that rules out a lot of potential counterexamples: if coproduct injections are monic in $\mathcal{C}^{op}$ but not in $\mathcal{C}$, then $\mathcal{C}$ admits no closed monoidal structure. To show this, suppose a coproduct injection $A\to A\coprod B$ coequalizes two parallel arrows $P\to A$, and apply the functor $[-,A]$ to everything, where $[-,-]$ is the internal hom of the closed monoidal structure. $\endgroup$ – Eric Wofsey Jun 4 '15 at 0:13
  • 2
    $\begingroup$ Since coproduct injections are monic in $\mathcal{C}^{op}$, $[A\coprod B,A]=[A,A]\times[B,A]\to [A,A]$ is epic, so the two maps $[A,A]\to [P,A]$ are equal. But this implies the original maps $P\to A$ were equal, since they are adjoint to the composition of the maps $[A,A]\to[P,A]$ with the map $1\to[A,A]$ that is adjoint to the identity on $A$ (where $1$ is the monoidal unit). $\endgroup$ – Eric Wofsey Jun 4 '15 at 0:15

Let me recall that injectivity of coproduct's injections follows from the distributivity of products over coproducts (rather than full extensivity of coproducts). Since every cartesian closed category is (obviously) distributive we cannot find counterexamples in cartesian closed categories.

However, the proof of injectivity of coproduct's injections highly relies on the cartesian structure of $\times$, and one should not expect to carry it to the context where product $\times$ is substituted by a general tensor $\otimes$.

One class of categories which cannot be cartesian closed (unless degenerated) are self-dual categories. I claim that very many of such categories do not have injective coproduct's injections, and this fact is (almost) unrelated to the existence of any closed monoidal structure. Here is an explicit example.

There are various notions of Chu spaces, but the underlying idea is common --- a Chu space is thought of as a "non-standard relation", and morphisms of Chu spaces are thought of as "adjoint pairs" between relations. Let me describe the category that is usually denoted by $\mathit{Chu}(\mathbf{Set}, \Omega)$. Its objects consist of typed binary relations: $$A = \langle A_!, A^*, A_! \times A^* \overset{A}\rightarrow \Omega \rangle$$ in $\mathbf{Set}$, and its morphisms $A \rightarrow B$ consist of pairs of functions (notice opposite directions!): $$f = \langle f_! \colon A_! \rightarrow B_!, f^* \colon B^* \rightarrow A^* \rangle$$ in $\mathbf{Set}$ that satisfy the following adjoint-like condition: $$B(b, f^*(a)) = A(f_!(b), a)$$ One may easily check that this category is self-dual, where the dualization swaps the domain with the codomain of a relation: $$(A^\bot)_! = A^*$$ $$(A^\bot)^* = A_!$$ $$(A^\bot)(b, a) = A(a, b)$$ and complete (thus, also cocomplete) --- limits are constructed point-wise: the first component of a Chu space inherits limits from $\mathbf{Set}$, and the second from $\mathbf{Set}^{op}$.

Like in many self-dual categories, objects may have many (co)global coelements --- i.e. there are non-trivial morphisms to the initial object $0 = \langle \emptyset, \{ {*} \}, \emptyset \rangle$ (just pick any Chu space $A$ whose $A_! = \emptyset$ and whose $A^*$ is non-trivial). Therefore, the unique morphism $0 \rightarrow 1$ is not mono. So the canonical coproduct's injection $0 \rightarrow 0 \sqcup 1 \approx 1$ is not a monomorphism in $\mathit{Chu}(\mathbf{Set}, \Omega)$.

There is a closed monoidal structure on $\mathit{Chu}(\mathbf{Set}, \Omega)$ given by the following tensor: $$(A \otimes B)_! = A_! \times B_!$$ $$(A \otimes B)^* = \{\langle h \colon A_! \rightarrow B^*, k \colon B_! \rightarrow A^* \rangle \colon B(b, h(a)) = A(a, k(b))\}$$ $$(A \otimes B)(a, b, h, k) = B(b, h(a)) = A(a, k(b))$$ In fact, $\mathit{Chu}(\mathbf{Set}, \Omega)$ is $\star$-autonomous (with linear implication $A \multimap B = (A \otimes B^\bot)^\bot$).

In particular, the construction is valid for any $\Omega$ (not necessarily the subobject classifier), and taking $\Omega = 1$ yields category $\mathbf{Set} \times \mathbf{Set}^{op}$ with linear exponentiation: $$\langle A_!, A^*\rangle \multimap \langle B_!, B^*\rangle = \langle {B_!}^{A_!} \times {A^*}^{B^*}, A_! \times B^*\rangle$$

| cite | improve this answer | |
  • 2
    $\begingroup$ Very good; +1. By the way, I wasn't explicitly aware of the fact that distributivity of products over coproducts implies monicity of coproduct inclusions. It's not very hard to prove though, if one remembers how to prove that distributivity implies that initial objects are strict (using a retraction argument). I have written up a proof here in case people want to see it: ncatlab.org/toddtrimble/published/… $\endgroup$ – Todd Trimble Jun 7 '15 at 15:24
  • 1
    $\begingroup$ @ToddTrimble Thanks! Why not put that at ncatlab.org/nlab/show/distributive+category ? $\endgroup$ – Mike Shulman Jun 8 '15 at 17:06
  • 1
    $\begingroup$ This is a nice example, thank you. I like $\mathbf{Set}\times \mathbf{Set}^{\mathrm{op}}$ even better as a simpler example. $\endgroup$ – Mike Shulman Jun 8 '15 at 17:17
  • $\begingroup$ @MikeShulman I've now done so (and redirected your link from the coprojection article). $\endgroup$ – Todd Trimble Jun 8 '15 at 17:52

I am following up on my comment.

The category of algebras over a commutative algebraic theory is monoidal closed. This is explained here http://ncatlab.org/nlab/show/commutative+algebraic+theory.

One can "commutativize" any theory by adding axioms expressing commutativity of any of its two operations (as in the nlab article). So, let us try to find a counterexample among categories of algebraic theories first, and then try to commutativize them.

Here is an example of a theory whose category of algebras may have non-injective coproduct injections: Take a theory which has $n$-ary terms $f, g, l$, $m$-ary terms $h$, $k$, and a binary term $p$, which satisfy

$$f(x_1, ..., x_n) = p(l(x_1, ..., x_n), h(y_1, ..., y_m))$$ $$g(x_1, ..., x_n) = p(l(x_1, ..., x_n), k(y_1, ..., y_m)).$$

Take a free algebra in this theory $F\{a_1, ..., a_n\}$ (shortly $F\{a\}$) on a set $\{a_1, ..., a_n\}$. Take a quotient of a free algebra $F\{b_1, ..., b_m\}/h(b_1, ..., b_m) \sim k(b_1, ..., b_m)$. In the coproduct of these two algebras we have

$$f(a) \sim p(l(a), h(b)) \sim p(l(a), k(b)) \sim g(a).$$

Thus, the coproduct injection from $F\{a\}$ is not injective (unless for some reason f(x) and g(x) equal to each other in the theory itself).

The theory of rings falls under this example by taking $f, g, l, h, k$ all the following zero terms $f = 2, g = 0, l = 1, h = 2, k = 0$, and $p(x, y) = xy$.

If we try to commutativize the theory of rings the constants get identified. In particular $2$ and $0$ get identified in the original theory, and we stay without a counterexample :).

In fact commutativization destroys the counterexample if the theory has any constant. Because, for a constant $c$, we should have $h(c, c, ...) = c = k(c, c, ...)$ by commutativity, and this forces $f(x)$ to equal $g(x)$ in the theory itself. This should be like this too, because the category of algebras of a commutative theory with a constant (necessarily unique) has a zero object, and thus has injective coproduct injections.

However, in other situations commutativization should be a harmless process. For example one can take a theory given by unary operation $f, g, l, h, k$ and a binary operation $p$, with the only axioms the above given two equations. Commutativazation of this theory should give a counterexample to the question.

| cite | improve this answer | |
  • $\begingroup$ I don't understand your argument about the coproduct. There are two copies of $a$ and $b$ in the coproduct, one coming from each summand. Which ones are you talking about? $\endgroup$ – Mike Shulman Jun 6 '15 at 23:14
  • $\begingroup$ I fixed it. For the counterexample one can assume f, g, l, h, k to be unary. I wrote in the arbitrary arity to include the analogy with the rings. $\endgroup$ – Dimitri Chikhladze Jun 6 '15 at 23:20
  • 1
    $\begingroup$ Thanks! Now I see. I believe your example probably works, but I'm accepting Michal's instead because I find it more intuitive and less ad hoc (YMMV). If I could accept more than one answer I'd accept yours too. $\endgroup$ – Mike Shulman Jun 8 '15 at 17:19
  • $\begingroup$ No worries. Admittedly, the example is artificial. But, if you try to find a universal algebraic example of non-injective coproduct injections, the argument which I used comes quite naturally, especially when thinking about the rings example (imho). $\endgroup$ – Dimitri Chikhladze Jun 8 '15 at 17:40
  • $\begingroup$ Dimitri, if you have a moment, please, write up your example in full detail --- just for the sake of having a reference to the construction (I suspect it may be a good instructional example). $\endgroup$ – Michal R. Przybylek Jun 9 '15 at 17:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.