Unitors and projections in cartesian category

Question

In a cartesian monoidal category we have the product with two projections $\pi_1$ and $\pi_2$, and the terminal object $1$. We also have unitors $\rho_A \colon A \times 1 \to A$ and $\lambda_A \colon 1 \times A \to A$. The obvious unitors are given by projections: $\rho = \pi_1$ and $\lambda = \pi_2$. Is it possible to have, for the same product and terminal object, a different set of unitors (and, by analogy, associators)? Would it still be called a cartesian monoidal category?

Note: Unitors determine projections. Given $! \colon A \to 1$ the unique morphism to the terminal object, we can define $\pi'_1: A \times B \to A$ as $\pi'_1 = \rho \circ (A \times \, !)$. But is this the same projection that defines the product?

Answer 1

First, let's look at the definition of a cartesian monoidal category on the nLab (emphasis mine):

Def. A cartesian monoidal category (usually just called a cartesian category), is a monoidal category whose monoidal structure is given by the category-theoretic product (and so whose unit is a terminal object).

So to unpack this a bit: a cartesian monoidal category $\mathcal C$:

• has binary products, i.e. for every two objects $A$ and $B$, we get a product $A \times B$, projections $\pi_1 : A \times B \to A$ and $\pi_2 : A \times B \to B$ such that the natural (w.r.t. $X$) transformation $Hom(X, A \times B) \to Hom(X, A) \times Hom(X, B) : \varphi \mapsto (\pi_1 \circ \varphi, \pi_2 \circ \varphi)$ has an inverse $\langle -,- \rangle$.

  From this, we can derive functoriality of $- \times -$ and naturality of $\pi_1$, $\pi_2$ and $\langle -,- \rangle$ w.r.t. $A$ and $B$.

• has a terminal object $\top$ such that for all $X$, $Hom(X, \top)$ is a singleton.

• is equipped with the following monoidal structure:

  • A monoidal product $A \otimes B := A \times B$,
  • A monoidal unit $I := \top$,
  • A right unitor $\rho := \pi_1 : A \times \top \cong A$,
  • A left unitor $\lambda := \pi_2 : \top \times A \cong A$,
  • An associator $\alpha := \langle \pi_1 \pi_1, \langle \pi_2 \pi_1, \pi_2 \rangle \rangle : (A \times B) \times C \cong A \times (B \times C)$.

So we can answer one of the OP's questions:

Q2: Would it still be called a cartesian monoidal category?

A2: No, by definition, the choice of unitors and associator in a cartesian monoidal category is fixed, once the cartesian structure (which is unique up to isomorphism) has been chosen.

The preceding question, however, can be interpreted in two subtly different ways:

Q1: Is it possible to have, for the same product and terminal object, a different(†) set of unitors and associator?

We can interpret this as either:

Q1a: Is it possible to have, for the same choice of $-\times -$ and $\top$, a different(†) set of unitors and associator?

Q1b: Is it possible to have, for the same choice of $-\times -$, $\pi_1$, $\pi_2$ (hence $\langle -,-\rangle$) and $\top$, a different(†) set of unitors and associator?

(†) Each time, "different" means different from the canonical ones chosen in the definition of a cartesian monoidal category.

A1a Interestingly, Q1a is subtly meaningless. Indeed, the canonical unitors are all defined in terms of $\pi_1$, $\pi_2$ and $\langle -,-\rangle$. So if you're not committing to a precise choice of projections and pairing, then the canonical unitors and associator are not yet defined and we cannot speak about being different from them.

@jean-baptiste-vienney gives an answer to the subtly meaningless question Q1a, and asking whether that answer is pertinent, is subtly equally meaningless. Indeed, it follows from the universal property of the binary product (consisting of both the object and the cone) that all binary products of the same two objects will be isomorphic cones. As such, the choice of binary products is unique up to isomorphism. In fact, this isomorphism can even be an automorphism, resulting in a different product cone with the same tip. This is what we see in @jean-baptiste-vienney's answer: they make a different but automorphic choice of cartesian products on the category of vector spaces, and the automorphism to the usual choice is given by multiplication by $2$. However, since different (though automorphic) cartesian products are chosen, we also get different canonical unitors and associator, which are compatible with the usual ones along the automorphism. Thus, while this different choice of cartesian products leads to different unitors and associator, these are still the canonical ones for this non-standard choice of products!

A1b: We have the following nearly-tautological theorem:

Thm. Let $\mathcal C$ be a category with cartesian products (i.e. $- \times -$, $\pi_1$, $\pi_2$ and $\langle -,-\rangle$) and a terminal object $\top$ equipped with a monoidal structure $(-\times-, \top, \lambda, \rho, \alpha)$. Then $\mathcal C$ is cartesian monoidal (i.e. $\lambda$, $\rho$ and $\alpha$ are the canonical ones) if and only if all three of the following properties are satisfied:

• $\lambda$ is compatible with $\pi_2$, in the sense that $\lambda = \pi_2$,
• $\rho$ is compatible with $\pi_1$, in the sense that $\rho = \pi_1$,
• $\alpha$ is compatible with projections in the sense that
  • $\pi_1 \alpha = \pi_1 \pi_1$,
  • $\pi_1 \pi_2 \alpha = \pi_2 \pi_1$,
  • $\pi_2 \pi_2 \alpha = \pi_2$.

Proof: This follows immediately from the universal property of the binary product. $\square$

This does not fully answer Q1b in the sense that I do not make any statements about the existence of unitors and associator that would be incompatible with projections, but it does show that simply asking for sensible interaction with the projections is enough to completely constrain the monoidal structure to be the canonical one.

Answer 2

It is possible for the same products and terminal object to have a different set of unitors i.e. unitors which are not given by the projections of these products.

Let $k$ be a field where $2 \neq 0$. And let $\mathcal{C}_0=Vec_k$. The vector space $A \times B$ is a product of $A$ and $B$ with projections given by $\pi^1_{A,B}(a,b)=a$ and $\pi^2_{A,B}(a,b)=b$. If $f:X \rightarrow A$ and $g:X \rightarrow B$ are two linear maps, then the unique map $\langle f,g \rangle : X \to A \times B$ such that $\pi^1_{A,B} \circ \langle f,g \rangle=f$ and $\pi^2_{A,B} \circ \langle f,g \rangle=g$ is given by $\langle f,g \rangle(x)=(f(x),g(x))$.

The vector space $A \times B$ can be made into a product of $A$ and $B$ with other projections. Define $\pi^{1'}_{A,B}(a,b)=2a$ and $\pi^{2'}_{A,B}(a,b)=2b$. It makes $A \times B$ into a product of $A$ and $B$. Let $f:X \rightarrow A$ and $g:X \rightarrow B$ be linear maps. The unique map $\langle f,g \rangle':X \rightarrow A \times B$ such that $\pi^{1'}_{A,B} \circ \langle f,g \rangle'=f$ and $\pi^{2'}_{A,B} \circ \langle f,g \rangle'=g$ is given by $\langle f,g \rangle' (x)=(\frac{1}{2}f(x),\frac{1}{2}g(x))$.

We will now proceed to give a precise definition of the notion of cartesian monoidal category. Let $\mathcal{C}$ be category. Suppose given a product cone $(A \times B,p_{A,B}:A \times B \rightarrow A, q_{A,B}:A \times B \rightarrow B)$ for every pair of objects $A,B \in \mathcal{C}$. We obtain a bifunctor $- \times -: \mathcal{C} \times \mathcal{C} \rightarrow \mathcal{C}$ by defining $f \times g=\langle p_{A,B};f, q_{A,B};g\rangle$ for every $f:A \rightarrow C$ and $g:B \rightarrow D$. We can also check that if we define

• $\lambda_A=q_{\top,A}:\top \times A \rightarrow A$,
• $\rho_A=p_{A,\top}:A \times \top \rightarrow A$,
• $\alpha_{A,B,C}=\langle p_{A \times B,C};p_{A,B}, \langle p_{A \times B,C};q_{A,B},q_{A \times B,C}\rangle\rangle:(A \times B) \times C \rightarrow A \times (B \times C)$

then, $\lambda,\rho,\alpha$ are natural transformations. Finally, we can check that $(\mathcal{C},\times,\top,\alpha,\lambda,\rho)$ is a monoidal category. Let's call cartesian monoidal category a monoidal category obtained in this way.

Applying this construction to the products cones $(A \times B,\pi^{1'}_{A.B},\pi^{2'}_{A,B})$ in $\mathcal{C}_0$ we obtain a monoidal category with:

• tensor product of morphisms $(f \times g)(x,y)=(f(x),g(y))$,
• associator $\alpha:(A \times B) \times C \rightarrow A \times (B \times C)$ given by $\alpha(x,y,z)=(2x,y,\frac{1}{2}z)$,
• unitor $\lambda:A \times 0 \rightarrow A$ given by $\lambda(a,0)=2a$,
• unitor $\rho:0 \times A \rightarrow A$ given by $\rho(0,a)=2a$.

Therefore the category $\mathcal{C}_0=\mathcal{Vec}_k$ is a monoidal category with tensor product given by $A \times B$ and monoidal unit $0$ for two different sets of unitors (and associators).

According to the definition given here of a cartesian monoidal category, which is the same as in Categories for the Working Mathematician, it is still a cartesian monoidal category.

Instead of believing that the construction of a cartesian monoidal category works well, the reader can alternatively check that the three diagrams for a monoidal category commute when defining unitors and associators as above.

Answer 3

I'm not 100% sure about this argument, but here goes...

First of all, $\rho_A$ has to be an isomorphism, so we should define it along with its inverse.

Second, $A$ is a product of itself and the terminal object, with projections $\pi_0 = 1_A$ and $\pi_1 = !_A$. That is, $(A, 1_A, !_A)$ is a product $A \times I$.

Finally, if $P$ and $P'$ are products of $A$ and $B$, then there is a unique isomorphism between them.

So it must be that $\rho_A$ is the unique isomorphism between $A \otimes I$ and $A$, because it's the only choice.

From here we can verify that $\rho = \pi_0$ by drawing the product diagram with $A$ as the product object and $A \otimes I$, $\pi_0$, and $ !_{A \otimes I}$ as the arbitrary object and morphisms.

Answer 4

When you talk about 𝐴×1, it must be one specific product, since it's defined up to an isomorphism (and the same holds about 1).

Of course there can be a bunch of isomorphisms A → A, giving you new unitors, unless you want some extra properties. Which ones?