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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?

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  • $\begingroup$ A product is both the object and the projections, so once you fix a particular categorical product object as the tensor product, then you also fix the projections - so would it be fair to rephrase this question as follows? Say $(P, \pi_0, \pi_1)$, is a product of $A$ and $B$, is there another product $(P, \pi'_0, \pi'_1)$ with $\pi_0 \neq \pi'_0$ or $\pi_1 \neq \pi'_1$ In other words, once you pick the object, do the projections become unique? $\endgroup$ – statusfailed Oct 28 at 19:14
  • $\begingroup$ It's not even clear that $\pi'_1$ and $\pi'_2$ define a product. They may not satisfy the universal condition. $\endgroup$ – Bartosz Milewski Oct 29 at 1:40
  • $\begingroup$ Let me add some motivation. In string diagrams we ignore unitors. So, for instance, if we have a morphism $f \colon A \to B$, the same diagram describes $f \circ \lambda_A$ and $\lambda_B \circ (id_1 \times f)$. I can prove this from the universal construction that defines $id_1 \times f$ only if the unitors are equal to the corresponding projections. $\endgroup$ – Bartosz Milewski Oct 29 at 20:07
  • $\begingroup$ I'm still having a bit of trouble interpreting this question, because I think it might be ill-posed as written. In particular: If for each object $A$ and $B$ you choose a specific product object $A \otimes B$ - which you must, because $\otimes$ has to be a functor - then you are asking if that object can be a product in more than one way. That is, can the object have multiple sets of valid projections. Is that right? (My problem with your wording is when you say "is it possible to have for the same product a different set of unitors - when you fix the product, you fix $\pi_1$.) $\endgroup$ – statusfailed Oct 30 at 21:11
  • $\begingroup$ Scratch that, I misunderstood! $\endgroup$ – statusfailed Oct 30 at 21:32
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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?

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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.

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  • $\begingroup$ I see what you mean. You actually have to prove that $(A, id_A, !_A)$ is a product. But that can be done by showing that it's universal among all $(X, f \colon X \to A, g \colon X \to 1)$. $g$ obviously must be $!$, and the commuting condition tells you that the unique morphism $h \colon X \to A$ must be $f$. Therefore $\rho_A$ is the unique isomorphism between $A \times 1$ and $A$. There is still one problem: The isomorphism may be unique, but there may be many morphisms $A \times 1$ to $A$ and $\pi_0$ may be just one of them. The diagram you're talking about has $\pi_0$ as the unique factor. $\endgroup$ – Bartosz Milewski Oct 30 at 23:11
  • $\begingroup$ It might be the case that there are many morphisms $A \times I \rightarrow A$, but only one of them satisfies the definition of a unitor, and that's $\pi_0$ - does this address the second part of your comment or did I miss the point? $\endgroup$ – statusfailed Oct 30 at 23:18
  • $\begingroup$ I think it boils down to: Can you show that $\pi_0 \colon A \times I \to A$ has an inverse? $\endgroup$ – Bartosz Milewski Oct 30 at 23:46
  • $\begingroup$ Not sure I agree - surely if you define the unitors as the unique isomorphism between $A \times I$ and $A$, then this implies that $\pi_0 = \rho$, so proving the existence of an inverse is unnecessary. $\endgroup$ – statusfailed Nov 1 at 13:08

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