In https://www.jmilne.org/math/xnotes/tc2018.pdf, page $7$ under the chapter on "Invertible objects" we call an object $L$ in a tensor category $(\mathcal{C},\otimes)$ (I will abbreviate this to $\mathcal{C}$) invertible if there is an equivalence of categories $X \leadsto L \otimes X$.

It is said that if there exists an object $L'$ in $\mathcal{C}$ such that $L \otimes L' \cong \mathbf{1}$, where $\mathbf{1}$ is the (unique up to isomorphism) unit object, then $L$ is invertible.

I tried showing this myself. I believe I can show that the functors $(L \otimes -)$ and $(- \otimes L')$ acts as (quasi)-inverses on arbitrary objects $X \in \mathcal{C}$. For example, we have

$$(L \otimes -) \circ (- \otimes L')(X) = L \otimes (X \otimes L') \overset{\text{id}_{L} \otimes \psi_{X,L'}}{\xrightarrow{}} L \otimes (L' \otimes X) \overset{\phi_{L,L',X}}{\xrightarrow{}} (L \otimes L') \otimes X \overset{\delta \otimes \text{id}_{X}}{\xrightarrow{}} \mathbf{1} \otimes X \overset{l_{X}^{-1}}{\xrightarrow{}} X$$ which is a composition of isomorphisms, hence an isomorphism. Without having written it out in full, showing that $(- \otimes L') \circ (L \otimes -)$ acts as the identity should involve basically the same isomorphisms.

However, I don't immediately see how to show that $(L \otimes -)$ and $(- \otimes L')$ composed, acts (up to isomorphism) as the identity on arbitrary morphisms $f:X \to Y$. I'd appreciate any hint/explanation.

To clarify, I want to show that if $L \otimes L' \cong \mathbf{1}$ for some object $L'$ in $\mathcal{C}$, then $X \leadsto L \otimes X$ defines an equivalence of categories.

Since functors that are faithful, full and essentially surjective defines an equivalence of categories, we could instead perhaps go by this route.

Since $L \otimes X \otimes L' \cong X$ for arbitrary $X$, we see that $X \leadsto L \otimes X$ is essentially surjective.

If $\text{id}_L \otimes f = \text{id}_L \otimes g$, then it is not immediately obvious that $f = g$. Without additional assumptions, I believe this clearly fails (thinking about the category $R\textbf{-Mod}$). Possibly one can use $L \otimes L' \cong \mathbf{1}$ here.

With regards to fullness, I am also not sure.

  • 1
    $\begingroup$ I believe the definition in terms of the isomorphism $L \otimes L' \cong \mathbf{1}$ is missing the triangle identities. That is, it's not a coherent definition. Your proof for objects should go through for morphisms if you use the coherences corresponding to the natural transformations you used for the object proof. $\endgroup$
    – S.C.
    Apr 19 at 3:25
  • $\begingroup$ Hm, I am not sure exactly how you mean, since $\delta$ is not in any of the coherence conditions. I also don't know what you mean by "the isomorphism $L \otimes L' \cong \mathbf{1}$ is missing the triangle identities". The link talks about dual objects $X^{\vee}$, but I don't see the relevance here. Or well, right, there might be some relevance, since if $X$ is invertible then $X$ is reflexive, i.e. $i_X:X \to X^{\vee \vee}$ is an isomorphism. I don't know, to be honest. $\endgroup$
    – Ben123
    Apr 19 at 3:40
  • $\begingroup$ The typical definition of an invertible object is that it's a dualizable one where the relevant maps are isomorphisms. Do you see how the triangle identities on the linked page apply to $\delta$? $\endgroup$
    – S.C.
    Apr 19 at 4:18
  • $\begingroup$ To amplify what @S.C. wrote: For an object $L$ to be invertible, it is not enough that there is some object $L'$ such that $L \otimes L'$ happens to be isomorphic to $\mathbb{1}$, just as in a monoid for an element $x$ to be invertible it is not enough that there is an element $y$ such that $xy = e$. $\endgroup$ Apr 27 at 20:19
  • $\begingroup$ @IngoBlechschmidt Hm, but that is what Deligne/Milne says, is it not? And I believe I managed to show this (recall that we are in tensor category here, so we have access to coherence conditions + commutator and associator). You can see my post about this on MO. $\endgroup$
    – Ben123
    Apr 27 at 23:15


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