\[ \gdef\EE{\mathbb{E}} \gdef\colim{\operatorname{colim}} \gdef\tensor{\otimes} \gdef\ZZ{\mathbb{Z}} \gdef\SSS{\mathbb{S}} \gdef\BB{\mathbf{B}} \gdef\CAlg{\operatorname{CAlg}} \gdef\Map{\operatorname{Map}} \gdef\map{\operatorname{map}} \gdef\End{\operatorname{End}} \gdef\Mod{\mathrm{Mod}} \gdef\PrL{\mathrm{Pr}^{\mathrm{L}}} \gdef\tel{\operatorname{tel}} \] This short post discusses a recent confusion about inverting things in a symmetric monoidal (\(\infty\)-)category, or the degenerate case, an "element" of an \(\EE_\infty\)-monoid.
Earlier today, I wrote quite a few paragraphs in the Algebraic Topology discord server about it, and rubberducked myself into the answer. I apologize if I wasted anyone else's time.
EDIT 2026-01-18: An error on how far we could generalise the setting in the Remark. EDIT 2026-02-18: Reworked some storytelling.
In the introduction, I'll be vague about the precise meaning of "localisation" and "to invert". I will mention a few concrete examples, but the reader is invited to come up with their own dialect of English where sense can be made.
If you have a family \(S\) of *s*tuff (that's a technical term) that you want to invert, you can make up a notion of \(S\)-local objects, as those who "treat things in \(S\) as if they are invertible", in an appropriate sense. In nice situations, each object has a universal approximation by an \(S\)-local object, and this approximation process is called localisation. It is out there for general reasons, which means we don't know a priori what they're built out of, except that they "solve the correct problem".
The stereotypical example is to "localise at a prime \(p\)", in which case \(S\) consists of all prime numbers except \(p\). Usually I'd be talking about rings or abelian groups, and we know exactly what localisations look like here: you can just write down fractions and routinely check that it works. You've seen this during your undergrads.
It takes some work to figure out what "fractions" means in other contexts, and they're not always the correct thing to do.
What's wrong with fractions? Can't you just…
Of course you can! It appears quite often and just works in a lot of situations, which is why people talk about it a lot. It's so commonplace that I nearly stopped thinking about it, and had to think to remember those cases where it famously doesn't work.
Here are a few common localisations, see if you spot any issue.
- Localisation in discrete commutative rings is given by fractions.
- Localisations of associative rings are given by fraction under some form of Ore's condition. In general they are build out of words
- Localising the category of chain complexes at the quasi-isomorphisms. There's a model using sequences of zig-zags, and if the stars align, a single zig-zag suffices.
- Inverting elements in ring spectra \(A\), or \(v_n\)-self maps of finite spectra.
- Inverting an object \(x\) in a symmetric monoidal category \(C\). This is often spoken about in motivic contexts, where one inverts the Tate motive (pointed projective line, an "twisted" version of \(S^2\)).
- Group completing the core of a symmetric monoidal category to obtain its "\(K\)-theory".
In analogy of fractions is the "telescope" construction on an endomorphism \(f : A \to A\) \[ \colim (A \xrightarrow{f} A \to \ldots) \eqqcolon \tel_f(A). \] More generally, given a map \(f : 1 \to M : C\), we can form the telescope \[ \colim (A \xrightarrow{f \tensor 1} M \tensor A \xrightarrow{f \tensor 1} M^{\tensor 2} \tensor A \to \ldots) \eqqcolon \tel_f(A). \] (There's also the variant of inverting \(X \to TX\) for an endofunctor \(T\) and other ones I don't know of.)
Notoriously in the point 3. and 4., the telescope in general does not suffice to invert, even though everything does seem commutative! The issue with group completion of spaces is no stranger for those who knows higher algebraic \(K\)-theory in its classical form: The telescope isn't even guaranteed to be an \(H\)-space, as there are some issues with the fundamental group actions. One must seeks out the aid of an arcane art — an ancient blessing, since lost to time, known as the \(+\)-construction of D. G. Quillen. (Apparently the construction was introduced by Kervaire, which Quillen applied to \(\BB\mathrm{GL}\) and made famous.)
I came to realise the presence of the issue only when I started learning about things in the motivic realm: I know some \(K\)-theory and group completion. Everyone straight off made it clear that it's a big issue, use the so-and-so simplicial delooping construction and hardly talked about the telescopes ever after. In algebraic contexts, or stably, one can compute its effect on homotopy groups, and it "just works".
Recently, I went back to reflect on this topic and realised the missing of some pieces. Way too often have I seen an author just telescope and say no second word about it, and I obviously hadn't been careful enough to question my fossilised algebraic instinct, spoiled by my training in set-based mathematics.
The answer lies not far beyond reach: It's already well-established in the motivic literature. More precisely, in the literature on the construction of the category of motivic spectra, where Robalo–Voevodsky–et al.'s criterion identifies a case where telescoping does invert, namely when the cyclic permutation \( (123) : x^{\tensor 3} \to x^{\tensor 3} \) is homotopic to the identity.
I suspect that the theory has been well known in the topology circle as well, since it's where the \(+\)-construction first appeared, exactly so that one can pass from telescope to group completion. It's also where we know that topological \(K\)-theory is actually given by a telescope. I've seen these during my studies, but not in an order so that I'd recognise it on the first glance. Now it's time to make up for it.
Why does it work, why does it not? What's going on?
Fix a (presentable) symmetric monoidal category \((C, \tensor, 1)\) and an \(\EE_\infty\)-algebra \(A :\CAlg(C)\). Inverting an \(x \in \End(A)\) is to find an initial algebra \(B\) under \(A\), on which \(x\) acts invertibly.
The problem of inverting \(x : C\) is then immediately reduced to the algebra case by viewing \(C : \CAlg(\PrL)\), with the endomorphism \((- \tensor x)\). We can moreover reduce to the case \(A = 1\) by passing to \(\Mod_A(C)\).
Reduced setup \((C, \tensor, 1)\) is a presentable symmetric monoidal category, \(f\) is an endomorphism of \(1\).
Warning I'm sweeping the size issues with \(\PrL\) under the rug. One might pretend there's a cardinality bound, go up a universe, or stick their head into the sand.
The localisations: What's going on
The full subcategory \(C_f\) of \(C\) on which \(f\) acts invertible is closed under all colimits and limits, so that the inclusion has both adjoints. Let \(L_f : C \twoheadrightarrow C_f\) denote the left adjoint.
It's also not hard to see that \(L_f\) is a symmetric monoidal localisation, i.e. the \(L_f\)-equivalences are closed under tensoring with any other object. We can now define \(E_f \coloneqq 1[f^{-1}] \coloneqq L_f(1)\). Now for any \(X : C\), the map \(X \to E_f \tensor X\) is an \(L_f\)-equivalence, since \(1 \to E_f\) is. \(E_f \tensor X\) is also \(L_f\)-local, as \(E_f\) is. In other words, the algebra \(E_f\) is the idempotent algebra which universally inverts \(f\), and the localisation is smashing, i.e.: \[ L_f(X) \simeq E_f \tensor X \]
As usual we let \(T_f \coloneqq \tel_f(1)\) be the sequential colimit as defined above. We note that \(1 \xrightarrow{f} 1\) is an \(L_f\)-equivalence, hence so is any \(X \to T_f \tensor X\). We can conclude that there is a unique map \(T_f \to E_f\) under \(1\), and that that they coincide if and only if \(T_f\) is already \(L_f\)-local \[ T_f \simeq E_f \Leftrightarrow f : T_f \to T_f \text{ is an isomorphism.} \]
Remark We can generalise to the case of an morphism \(f : 1 \to I\), and consider the full subcategory on objects \(X\) where \[ f \tensor 1 : X \to I \tensor X \text{ is an isomorphism.} \] In addition, we demand that \(I\) be
dualisable(ERRATUM: invertible!). This guarantees the following:
- \((- \tensor I)\) preserves limits, so that \(C_f\) is closed under limits. The left reflection \(L_f : C \twoheadrightarrow C_f\) exists.
- \(C_f\) is closed under powers, so that \(L_f\)-equivalences are closed under tensors. \[ \hom_C(Z, X) \to I \tensor \hom_C(Z, X) \simeq \hom_C(Z, I \tensor X). \] This makes \(L_f\) into a monoidal localisation.
- \(T \xrightarrow{f \tensor 1} I \tensor T\) is an \(L_f\)-equivalence. Indeed, \(f \tensor 1\) being an \(L_f\)-equivalence is the same as \(I^\vee \tensor X \xrightarrow{f^\vee \tensor 1} X\) being an equivalence for all \(X : C_f\), which follows if the coevaluation \(I^\vee \tensor I \to 1\) is an isomorphism.
For the first two points to hold, we only need dualisability of \(I\). The third is where we needed invertibility, in order to make \(T_f\) even plausible as a candidate for \(E_f\).
Example In \(C = \mathrm{An}\), we consider the algebra \(A = \mathrm{core}(\mathrm{FinSet})\), the free \(\EE_\infty\)-space on a point, with the distinguished endomorphism \((- \cup \{\ast\})\). The telescope construction is \[ T_f \simeq \ZZ \times \BB\Sigma_\infty \not\simeq E_f = \Omega^\infty \SSS \simeq \ZZ \times \BB\Sigma_{\infty +}. \] We can verify that \((- \cup \{\ast\})\) doesn't act invertibly: It includes each copy of \(\Sigma_\infty\) into the the next copy as the stabiliser of a point.
When it works, it works.
Let's first summarise our setup:
\((C, \tensor, 1)\) is a symmetric monoidal category, \(I : C\) is an invertible object and \(f: 1 \to I\) is a morphism.
\(C_f\) is the bireflective subcategory on which \(f\) acts invertibly. The left localisation \(L_f\) is a smashing localisation, given by an idempotent algebra \(E_f\).
We would like to know whether \(E_f\) is equivalent to the telescope \[ T_f \coloneqq \colim ( 1 \xrightarrow{f} I \xrightarrow{f \tensor 1} I^{\tensor 2} \to \ldots) \]
The motivic literature contains the following criterion for when the telescope does give the correct answer.
Criterion (Robalo–Voevodsky and many other people) If for some \(n \geq 2\), the cyclic permutation \[ (12\ldots n) : f^{\tensor n} \Rightarrow f^{\tensor n} : 1 \to I^{\tensor n} \] is homotopic to the constant homotopy, then \(T_f \simeq E_f\).
The idea It suffices to show that \(f\) acts invertibly on \(T_f\).
There's an obvious equivalence \(T_f \xrightarrow{\sim} T_f \otimes I\), given by a cofinal inclusion of the two colimit diagrams.
The issue is that this inclusion differs from the action of \(f\) on \(T_f\). An inclusion would have identity arrows and identity homotopies everywhere. The homotopies that are involved with \(T_f \xrightarrow{f \tensor 1} I \tensor T_f\) come from the bifunctoriality of \(- \tensor -\). If one thinks about it, the homotopy is exactly the cyclic permutation \((12)\) acting on \(f^{\tensor 2}\). If we skip terms and use the cofinal subdiagram \[ A \xrightarrow{f^{\tensor n}} I^{\tensor n} \tensor A \to \ldots \] then the difference is the permutation \((12\ldots (n+1))\). If this homotopy is constant, then the construction goes through, and the action of \(f\) on \(T_f\) can be identified with a cofinal inclusion, and is hence an isomorphism.
Corollary If \(C\) is stable, then \(T_f \simeq E_f\).
Proof The action of \(\Sigma_n\) on the above map is given by a map \[ \BB\Sigma_n \to \Map_C(1, I^{\tensor n}). \] When \(C\) is stable, the right-hand side is the 0-th space of spectrum, and the map factors as \[ \SSS[\BB\Sigma_n] \to \map_C(1, I^{\tensor n}). \] The cyclic permutation then acts through its image in \[ \Sigma_n = \pi_1\BB\Sigma_n \to \pi_1\SSS[\BB\Sigma_n]. \] The latter is, by Hurewicz theorem, \(\pi_1\SSS \oplus \Sigma_n^{\textrm{ab}} \simeq \ZZ/2[\eta] \oplus \{\pm\}\). The permutation lands in the second factor as its sign, and is \(+\) whenever \(n\) is odd.
Corollary The classifying spaces \(\mathrm{Vect}_{\mathbb{K}}\) of vector bundles (\(\mathbb{K} = \mathbb{R},\mathbb{C}\)) have group completion \(\mathrm{kgl}_{\mathbb{K}}\) (under Whitney sum) given by the corresponding telescopes: \[ \mathrm{ko} \coloneqq (\mathrm{Vect}_{\mathbb{R}})^{\textrm{grp}} \simeq \ZZ \times \BB\mathrm{O}, \mathrm{ku} \coloneqq (\mathrm{Vect}_{\mathbb{C}})^{\textrm{grp}} \simeq \ZZ \times \BB\mathrm{U}. \] In fact, the topological categories \(\mathrm{Vect}_{\mathbb{K}}^{\textrm{inj}}\) of finite dimention \(\mathbb{K}\)-vector spaces and injections "Picard complete" to categories \(\mathrm{kgl}_{\mathbb{K}}^{\textrm{inj}}\) through their telescopes. The mapping spaces are \[ \Map_{\mathrm{kgl}_{\mathbb{K}}^{\textrm{inj}}}(m,n) \simeq \begin{cases}\emptyset & m > n \\ \mathrm{GL}_\mathbb{K}(\infty)/\mathrm{GL}_\mathbb{K}(-m+n) & m \leq n\end{cases} \]
Proof The action of \((123)\) on \(\mathbb{R}^3\) is homotopic to the identity. In the complex case, even \((12)\) acts as identity.