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Let us take the following assumptions: $\mathscr{M}$ a monoidal category, $X,Y,Z$ three objects in the category, and $f: Y \to Z$ a morphism. If the morphism $$ \mathrm{id}_X \otimes f: X \otimes Y \to X \otimes Z $$ is an isomorphism then can we imply that $f$ is also an isomorphism?

Edit: Due to Daniel's comment below this does not seem to be true, so I wonder if there are some special type of categories where such problems do not arise. For example if the category is also abelian and the tensor product is linear in the obvious sense? Indeed if the category admits duals then this should be true?

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    $\begingroup$ 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. $\endgroup$
    – daniel gratzer
    Commented Sep 28, 2023 at 14:13
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    $\begingroup$ 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. $\endgroup$
    – daniel gratzer
    Commented Sep 28, 2023 at 14:20
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    $\begingroup$ Also if you collect your comments together into an answer I am happy to accept it. $\endgroup$
    – Didier de Montblazon
    Commented Sep 28, 2023 at 15:03
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    $\begingroup$ 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. $\endgroup$
    – Tobias Fritz
    Commented Sep 28, 2023 at 15:36
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    $\begingroup$ 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. $\endgroup$
    – Tobias Fritz
    Commented Sep 28, 2023 at 15:38

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Of course not. Take X=0 for example in the additive case. But a sufficient condition in the context of abelian tensor categories is that X is faithfully flat. The most well-known case here is the category of modules over a ring, in which case we get faithfully flat modules.

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