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Let $(\mathcal{C}, \otimes , 1)$ be a monoidal category, and let $\mathrm{End}_{\mathcal{C}} (1)$ be the ground monoid of $\mathcal{C}$ - which is a commutative monoid. If $r_X : X \otimes 1 \to X$ denotes the right unit constrain of $\mathcal{C}$, then $f \cdot \alpha := r_Y (f \otimes \alpha) r_X^{-1}$ defines a right action on $\hom_{\mathcal{C}} (X;Y)$.

Question 1. When is this action free? That is, under what conditions $f \cdot \alpha = f$ implies $\alpha = \mathrm{id}_1$?

Question 2. Is it possible to loosen the conditions on $\hom_{\mathcal{C}} (1;Y)$?

For instance, in the category of $k$-modules this action is not free even when $k$ is a field (see Maxime Ramzi's comment). When $k$ is a ring the situation is even worse because of torsion.

Edit. More concretely, I'm trying to give sufficient conditions to mimic the following situation in the category of $k$-modules ($k$ a field): if $0 \neq v \in V$ and $\lambda \in k$, then $v = \lambda v$ implies $\lambda =1$.

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  • $\begingroup$ The action isn't free in the case you describe because of $f=0$. I feel like it's going to be hard to isolate nice situations where this happens because of phenomena like that $\endgroup$
    – Maxime Ramzi
    Commented Apr 17, 2021 at 11:44
  • $\begingroup$ Nice catch! I removed that $\endgroup$
    – Minkowski
    Commented Apr 17, 2021 at 12:54
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    $\begingroup$ Small observation: in general, even the action of a monoid on itself isn't necessarily free in this sense. (Only if the monoid is cancellative.) So maybe a good first question is to ask when $\operatorname{End}(1)$ is cancellative? $\endgroup$
    – Achim Krause
    Commented Apr 17, 2021 at 14:29
  • $\begingroup$ Are you quantifying over $X$ and $Y$? That is, are you looking for conditions on $X,Y$ such that the ground monoid acts freely on $Hom(X,Y)$? Or are you looking for conditions such that if (for all $X,Y$, for all $f: X \to Y$, $\alpha \cdot f= f$), then $\alpha = 1$? If the latter, then note that when $\mathcal C$ is modules over a nonzero ring, freeness always fails when $X=0$ or $Y=0$, so whatever condition you formulate must rule out that case somehow. $\endgroup$
    – Tim Campion
    Commented Apr 17, 2021 at 20:40
  • $\begingroup$ What I'm really trying to mimic is the multiplication by the ground field in the category of $k$-modules (I've edited my question to that end). But looking at your comments I see not even $\mathsf{Mod}_k$ has a free action, so indeed I should be looking for something not that restrictive. $\endgroup$
    – Minkowski
    Commented Apr 18, 2021 at 10:16

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