# Invertible objects in tensor categories.

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.

• 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.
– S.C.
Apr 19 at 3:25
• 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. Apr 19 at 3:40
• 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 $\delta$?
– S.C.
Apr 19 at 4:18
• 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 $xy = e$. Apr 27 at 20:19
• @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. Apr 27 at 23:15