Timeline for Isomorphisms after tensoring with the identity in a monoidal category
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 2, 2023 at 5:20 | review | Close votes | |||
| Oct 7, 2023 at 3:03 | |||||
| Oct 1, 2023 at 21:30 | comment | added | LSpice | @MatthewNiemiro, re, but, as pointed out above, why would you want some exotic object like the rational numbers when you could just consider the trivial group? 😄 | |
| Oct 1, 2023 at 20:30 | comment | added | Matthew Niemiro | Rationalization? | |
| Oct 1, 2023 at 20:06 | answer | added | Martin Brandenburg | timeline score: 6 | |
| Sep 29, 2023 at 13:35 | comment | added | Didier de Montblazon | @TobiasFritz: Thanks a lot | |
| Sep 28, 2023 at 17:10 | comment | added | Tobias Fritz | Hence the constructed morphism is a left inverse of $f$. The other direction (right inverse) is analogous. | |
| Sep 28, 2023 at 17:08 | comment | added | Tobias Fritz | $(\beta \otimes \mathrm{id}_Y) \circ g \circ (\alpha \otimes \mathrm{id}_Z) \circ f = (\beta \otimes \mathrm{id}_Y) \circ g \circ (\mathrm{id}_X \otimes f) \circ (\alpha \otimes \mathrm{id}_Y) = (\beta \otimes \mathrm{id}_Y) \circ (\alpha \otimes \mathrm{id}_Y) = (\beta \circ \alpha) \otimes \mathrm{id}_Y = \mathrm{id}_Y$. | |
| Sep 28, 2023 at 17:07 | comment | added | Tobias Fritz | Sure. Suppose that $\alpha : I \to X$ and $\beta : X \to I$ compose to the identity, and that you're given an inverse $g : X \otimes Z \to X \otimes Y$ for your $\mathrm{id}_X \otimes f$. Then I claim that $(\beta \otimes \mathrm{id}_Y) \circ g \circ (\alpha \otimes \mathrm{id}_Z)$ is an inverse of $f$, where I'm assuming that $\mathscr{M}$ is strict so that I can omit the unitors. The fact that this is the inverse of $f$ is easiest to see in terms of string diagrams, but here's one direction in equational form: | |
| Sep 28, 2023 at 16:55 | comment | added | Didier de Montblazon | @TobiasFritz: Could you explain the comment " it is true in ℳ if there are morphisms 𝐼→𝑋→𝐼 that compose to the identity, where 𝐼 is the monoidal unit"? Or a suitable reference would also be great. Thank you! | |
| Sep 28, 2023 at 15:38 | comment | added | Tobias Fritz | Duals do not help: you can again take $X$ to be empty to get counterexamples in $\mathbf{Rel}$, the category of sets and relations with cartesian product as monoidal structure, which is compact closed. And similarly in the category of finite-dimensional vector spaces over a field, if you want both abelian and compact closed. | |
| Sep 28, 2023 at 15:36 | comment | added | Tobias Fritz | This is not even true in Set with cartesian product: as @danielgratzer says, just take $X$ to be empty :) However, it is true in $\mathscr{M}$ if there are morphisms $I \to X \to I$ that compose to the identity, where $I$ is the monoidal unit. This applies in many cases. | |
| Sep 28, 2023 at 15:03 | comment | added | Didier de Montblazon | Also if you collect your comments together into an answer I am happy to accept it. | |
| Sep 28, 2023 at 14:47 | comment | added | Didier de Montblazon | Sorry to keep "moving the goal post" but does $\mathbf{Ab}$ admit duals? | |
| Sep 28, 2023 at 14:20 | comment | added | daniel gratzer | I don't believe this is true even in $\mathbf{Ab}$. Consider tensoring with $\mathbb{Z}/p$ and some map $\mathbb{Z}/q \to \mathbb{Z}/q$ where $p$ and $q$ are distinct primes. The tensor product $\mathbb{Z}/p \otimes \mathbb{Z}/q$ is 0, so the induced map is always an isomorphism but obviously not every map $\mathbb{Z}/q \to \mathbb{Z}/q$ needs to be an isomorphism. | |
| Sep 28, 2023 at 14:17 | history | edited | Didier de Montblazon | CC BY-SA 4.0 |
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| Sep 28, 2023 at 14:16 | comment | added | Didier de Montblazon | @Daniel: Thanks a lot for the comment it is very helpful. Do you know of some special type of category where such problems do not arise. For example if the category is also abelian and the tensor product is linear in the obvious sense? | |
| Sep 28, 2023 at 14:13 | comment | added | daniel gratzer | No not generally. In fact, the theory of "smashing localizations" deals exactly with inverting maps which do become invertible after tensoring with some fixed object. Additionally, in many categories taking the cartesian product with 0 results in an isomorphism irrespective of f. | |
| Sep 28, 2023 at 14:00 | history | asked | Didier de Montblazon | CC BY-SA 4.0 |