Suppose we have a Quillen adjunction $L\colon {\mathcal C} \leftrightarrow {\mathcal D}: R$. For convenience let us assume that all objects of $\mathcal C$ are cofibrant and all objects of $\mathcal D$ are fibrant, so that $L$ and $R$ are homotopy functors. Let $(L', R')$ be another Quillen adjunction. Let's suppose that we have a natural transformation $\alpha_L\colon L \to L'$. It induces a natural transformation $\alpha_R\colon R' \to R$. Let us assume that $\alpha_L$ is a weak equivalence. If I am not mistaken, it follows that $\alpha_R$ is a weak equivalence.

Let $T$ and $T'$ be the monads $RL$ and $R'L'$ respectively. As far as I see, in general there is no map of monads between $T$ and $T'$.

Question : Are the homotopy categories of $T$ algebras and $T'$-algebras equivalent?

Remark : The adjunction $(L, R)$ induces a Quillen adjunction between $T$-algebras and $\mathcal D$. This adjunction is sometimes (quite often?) a Quillen equivalence. In this case one may say that $T$ satisfies homotopy descent. It seems that if $T$ satisfies homotopy descent, then so does $T'$, and in this case Ho$(T-\mathrm{alg})$ and Ho$(T'-\mathrm{alg})$ are equivalent, since both are equivalent to Ho$(\mathcal D)$. I want to know if there is a direct way to compare algebras over $T$ and $T'$, without using descent.


Hi Greg, I'll give an alternative, elementary, starting point towards an answer. Dylan, there is a question about using the bar construction that seems maybe to be missing from your answer, and I don't understand your statements about dealing with multiplicative structure or with uniqueness. Also, if you will forgive a little teasing, if your favorite tool is a sledgehammer, then every nail looks like a rock.

I'll use the two-sided bar construction from "The geometry of iterated loop spaces", [12] on my web page; see [112] for a modernized discussion and some category theory relevant to the question. For a $T$-algebra $A$, define a $T'$-algebra $F(A)$ by $F(A) = R'B(L,T,A)$, where $T$ acts on the right of $L$ via $\epsilon R$.
This is a $T'$-algebra since $R'D$ is a $T'$-algebra for any $D\in \mathcal D$. Similarly, For a $T'$-algebra $A'$, define a $T$-algebra $G(A')$ by $G(A') = RB(L',T',A')$. Of course, $F$ and $G$ are functors.

I assume that we have natural weak equivalences $$\gamma'\colon B(R'L, T, A) \to R'B(L,T,A)$$ and $$ \gamma\colon B(RL', T', A') \to RB(L',T',A').$$ We expect the second to come from a weak equivalence of the general form $|RX|\to R|X|$ for a simplicial object $X\in \mathcal C$, and similarly for the first. With $R = \Omega$ and $\mathcal C$ the category of based spaces, proving that there is such a weak equivalence $\gamma$ was the hardest thing technically in [12]. It is something like this that I find missing in your sketch, Dylan, but if we can see directly that $B(R'L, T, A)$ is a $T'$-algebra (say by commuting $T'$ past $B$) and similarly for $B(RL',T',A')$, then perhaps that is not needed here.

Since $T = RL$, we have the chain of natural weak equivalences (ignoring algebra structure) $$ A \leftarrow B(T,T,A) \leftarrow B(R'L,T,A) \to F(A) $$ where the middle arrow is induced by $\alpha\colon R'\to R$, which I'll assume given (rather than taking a conjugate). Similarly, we have the chain of weak equivalences $$ A' \leftarrow B(T',T',A') \rightarrow B(RL',T',A') \to G(A'). $$ Therefore $F(G(A'))\simeq A'$ and $G(F(A))\simeq A$. It remains to prove that these equivalences (or related ones) are equivalences of $T'$-algebras and $T$-algebras. I haven't had time to try.

  • $\begingroup$ Teasing is always welcomed. For the type of "uniqueness" I wanted, see the paragraph (and ensuing proofs if you like) beginning section 6.2.3 of Lurie's Higher Algebra. What I meant about the bar construction is just this: in the case given, algebras over a monad are generated by sifted (homotopy) colimits of free algebras. So if the given equivalence $T \rightarrow T'$ induces a map $T-alg \rightarrow T'-alg$ preserving sifted colimits then we can reduce to the free case. I should have made this assumption more explicit (I can't think of a counterexample to this property of an equivalence $\endgroup$ – Dylan Wilson Mar 19 '13 at 3:39
  • $\begingroup$ off the top of my head, now that I think about it.) Like you, I haven't had time to look at the multiplicativity statement. I could just be wrong! :) $\endgroup$ – Dylan Wilson Mar 19 '13 at 3:40
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    $\begingroup$ re sledgehammers: This is definitely laziness on my part. But I am somewhat okay with this... it's nice when the proof of some fact is as intuitive as the idea behind it. Now, both of our "proofs" (maybe mine's wrong...) had the same outline "prove it for free things, then do a little formalism to get everything else", we just went about it with slightly different language. $\endgroup$ – Dylan Wilson Mar 19 '13 at 3:45
  • $\begingroup$ Hi Peter. This is a helpful answer, thanks. It is interesting that the assumption that the map $L\to L'$ is an equivalence is not directly relevant to your argument. What matters is that $R$ and $R'$ can be moved past the bar constructions. The assumptions end up being equivalent to a form of Barr-Beck criterion for monadic descent. They guarantee that $T$ and $T'$-algebras embed into $\mathcal D$. You show that the images are the same. I had wondered whether the assumption that the maps $\alpha_L, \alpha_R$ are equivalences would give a more direct way to compare the monads. Perhaps not. $\endgroup$ – Gregory Arone Mar 19 '13 at 7:15

Unless I'm mistaken (very possible), the answer is "yes".

Here's the general idea: Every algebra may be obtained as the geometric realization of its bar complex. So we may reduce to the case of a free algebra, i.e. we need only construct some sort of natural weak equivalence $TX \rightarrow T'X$ preserving the multiplication structure. I don't know how to make precise this choice in a model category setting without using an ugly zig-zag and convoluted argument (but that is more likely due to ignorance than anything else.)

If we replace all model categories in sight with their underlying $\infty$-category (I'm ignoring some set-theoretic issues here, or assuming that the categories are simplicially enriched), then the above becomes an actual proof: The Quillen adjunction induces an adjunction of $\infty$-categories, and in this case a natural isomorphism of functors admits an $\infty$-categorical inverse. So we can define the map $$ T \rightarrow T' $$ as the composition $$\alpha_R^{-1}L' \circ R\alpha_L: RL \rightarrow RL' \rightarrow R'L'$$

(Here is where, in the model category case, I would have a zig-zag, and would have to do some sort of work... I don't actually know how the argument would go off the top of my head.)

This map respects the multiplication (the only proof I can think of uses that adjoint pairs are unique up to unique isomorphism, where "unique" in this setting means "parameterized by a trivial Kan complex". But it's possible this is more obvious than I think.)

It is a natural isomorphism, and so we're good.

I don't think we can get away with any other type of proof: you need some sort of "multiplication preserving zig-zag of weak equivalences of functors" argument to prove this at all, because if you provide some natural zig-zag of weak equivalences between $T$-algebras and $T'$-algebras then, in particular, you need to do it for free $T$-algebras and $T'$-algebras, and multiplication preserving weak-equivalences like this can only come in the form of some zig-zag of natural transformations of $T$ with $T'$ since that's where the algebra structure on free algebras comes from.

  • $\begingroup$ I should note that proving the result in the infinity-category language is strong enough (in good cases) to give the result about model categories. The above showed that the $\infty$-categories of $T$-algebras and $T'$-algebras are equivalent, and if both of these were, say, combinatorial model categories, then it would follow that the model categories are connected by a zig-zag of Quillen equivalences (I think you need only one zig and one zag, but I haven't convinced myself of this yet.) $\endgroup$ – Dylan Wilson Mar 18 '13 at 14:30
  • $\begingroup$ Thanks. I would be happy to see a little more details of the argument that the map that you constructed is multiplicative. Do you have a reference to the statement that "adjoint pairs are unique"? $\endgroup$ – Gregory Arone Mar 18 '13 at 19:23
  • $\begingroup$ I'll add in details for the map to be multiplicative a little later. For the uniqueness statement I guess I don't know a reference, but the proof is easy: Given $L$ I can recover $R$ by looking at the functor $\text{Hom}(L(-), (-))$ and vice versa. This statement makes sense in quasi-category land but now hom-sets are replaced by hom-spaces and "recover" has the usual "up to a contractible choice" caveat. $\endgroup$ – Dylan Wilson Mar 19 '13 at 0:33
  • $\begingroup$ Also- I'm starting to doubt whether $\alpha_R$ comes for free with $\alpha_L$, satisfying the conditions that we want. I think we want what MacLane calls "conjugate natural transformations." See Theorem 2, Section 7, Chapter IV or Categories for the Working Mathematician $\endgroup$ – Dylan Wilson Mar 19 '13 at 0:36

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