Let Mon be the category of topological monoids. I am happy to work with the model structure mentioned here:

Model Structure/Homotopy Pushouts in topological monoids?.

I'm looking for a reference for the following statement, which I believe to be true. Suppose $X$ and $Y$ are topological monoids, with $X$ cofibrant, and $Y$ group-like.

Assertion: The map of spaces

$$ \text{monoid-maps}(X,Y) \to \text{maps}_\ast(BX,BY) $$

is a weak homotopy equivalence.

My guess is that one could prove this by induction, using the fact that $X$ is a retract of an object given by attaching free things. For example, here is a verification of the statement when $X = FU$, the free monoid on the points of a based space $U$. In this case,

$$ \text{monoid-maps}(FU,Y) = \text{maps}_\ast(U,Y) , $$

whereas $$ \text{maps}_*(BFU,BY) \simeq \text{maps}_\ast(\Sigma U,BY) $$ using, say James theorem $FU \simeq \Omega\Sigma U$. Since $Y$ is group-like, we have $Y \simeq \Omega BY$ and we get

$$ \text{maps}_*(\Sigma U,BY) \simeq \text{maps}_\ast(U,Y) $$ verifying the assertion in this special case.

More generally, it seems to me that if $X = \text{colim}(X_0 \leftarrow FA \to FB)$ where $(B,A)$ is a cofibration pair (with the colimit taken in topological monoids), and if the assertion is true for $X_0$ then it is also true for $X$ using the above and by noting that (i) function spaces convert pushouts in the domain to pullbacks and (ii) the classifying space functor preserves homotopy pushouts. This would then give the inductive step.

Added Later: I'd like to reiterate that I'm really looking for a decent reference.

  • $\begingroup$ I hesitate to continue to add things, so I'll just leave this as a comment... But it just occurred to me that the following could work and would be more elementary: (i) The functor monoids--->spaces preserves and reflects geometric realizations of simplicial objects (I think you can prove this directly, it's the homotopy version of saying monoids are algebras for a monad), (ii) the natural map $\vert F^{\bullet}X\vert \rightarrow X$ from the free resolution of $X$ is a weak equivalence, (iii) B commutes with geometric realizations, (iv) you've already proven the result for free things. $\endgroup$ Apr 16, 2013 at 16:39
  • $\begingroup$ (The only place I can see potential trouble is some sort of cofibrancy issue. Sorry I am of limited use! I hope someone comes along and adds a decent answer.) $\endgroup$ Apr 16, 2013 at 16:40

1 Answer 1


EDIT 2: Sorry about the confusion, I will try to be careful now, and I'll put some comments at the end to clear up the business about adjoints etc. Here's what we're going to do.

We'd like to show that $$ \text{Map}_{\text{Mon}}(X, Y) \rightarrow \text{Map}_{Spaces_*}(BX, BY) $$ is an equivalence when $X$ is cofibrant and $Y$ is grouplike. The strategy will be to consider the string $$ \text{Map}_{\text{Mon}}(X,Y) \rightarrow \text{Map}_{\text{Spaces}_*}(BX,BY) \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY) \rightarrow $$

$$ \text{Map}_{\text{Mon}}(X, \Omega'BY) \cong \text{Map}_{\text{Mon}}(X,Y) $$ where $\Omega'$ denotes the Moore loop space (strictly associative multiplication), and show that the composite is homotopic to the identity. This clearly reduces down to showing that the third map is a weak equivalence, since I may as well have started the string with the inverse of the last equivalence ($\Omega' BY$ and $Y$ are interchangeable).

So we want to show that $$ \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY) \rightarrow \text{Map}_{\text{Mon}}(X, \Omega'BY) $$ is a weak equivalence when $X$ is cofibrant ($Y$ can be arbitrary).

We will prove this by showing that these two spaces represent the same functor on the homotopy category of spaces, the fundamental fact being that the natural map $\Omega'B Y \rightarrow \Omega'B\Omega'B Y$ is a weak equivalence.

Given a map $K \rightarrow \text{Map}_{\text{Mon}}(X, \Omega'BY)$ we get a map $K \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'B\Omega'BY)$, and composing with the natural weak equivalence gives us a map $K \rightarrow \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY)$. One can check that composition gives back the original map, up to homotopy. This proves surjectivity of

$$ \text{Map}_{h\text{Spaces}}(K, \text{Map}_{\text{Mon}}(\Omega'BX, \Omega'BY)) \rightarrow \text{Map}_{h\text{Spaces}}(K, \text{Map}_{\text{Mon}}(X, \Omega'BY)) $$ to get injectivity just note that we can recover the original value by applying $\Omega B'$ and using the weak equivalence again.

This completes the proof.

In my "second edition" I erroneously stated that there was a left adjoint to the inclusion of group-like topological monoids into monoids. Ricardo gracefully explains why this can't be true. However, it is a consequence of the above argument that there is a homotopical left adjoint, namely $\Omega'B$. (We've shown that the image of $\Omega'B$ is a homotopically reflective subcategory, and so all that's left is to note that the essential image is all group-like topological monoids.)

I've also erased the original answer, since it seems silly now that we have this one.

A reference for the argument I gave that basically showed that $\Omega'B$ was a localization functor can be found in Higher Topos Theory Proposition Lots more love can be found throughout Higher Algebra; one can, for example, deduce a similar statement about $\mathbb{E}_1$-spaces in general, which actually implies this result since the inclusion of monoids into $\mathbb{E}_1$-spaces is an equivalence of $\infty$-categories.

Hopefully this is all correct now! Let me know if there's still errors.

  • $\begingroup$ (Your cofibrancy condition is used to make sure that the mapping space in the actual model category agrees with the "derived" mapping space; sorry, I forgot to mention that.) $\endgroup$ Apr 14, 2013 at 16:37
  • 2
    $\begingroup$ @Dylan: your proof looks correct, but it seems way too high tech. Doesn't the sketch I provide above look somewhat simpler? Isn't there a sketch of this sort somewhere in the literature? Anyway, I'm happy though that you mentioned Lurie's equivalence: $\text{Monoids} \to \text{Mon}_{\Bbb E_1}$, which is also of use to me. $\endgroup$
    – John Klein
    Apr 14, 2013 at 17:30
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    $\begingroup$ @Dylan: I am afraid the inclusion of the full subcategory of group-like monoids in the category of monoids in topological spaces does not have a left adjoint. This is because a reflective subcategory of a complete category is complete, and the inclusion preserves limits. Now observe that there are limits of group-like monoids which are not group-like. For example, the pullback of $[0,+\infty) \to S^1 \leftarrow 1$ (where the map $[0,+\infty) \to S^1$ is given by complex exponential) in the category of topological monoids is isomorphic to $\mathbb{N}$, which is not group-like. $\endgroup$ Apr 14, 2013 at 21:43
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    $\begingroup$ @Dylan: I don't understand why you write $\text{Map}_{\text{Mon}}(X, \Omega'BY) \cong \text{Map}_{\text{Mon}}(X,Y)$. It's at best a weak equivalence. Secondly, why is the map $Y\to \Omega' BY$ a monoid map. It's an $A_\infty$-map, I agree, but I doubt it's a monoid map. Am I missing something? $\endgroup$
    – John Klein
    Apr 14, 2013 at 23:45
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    $\begingroup$ @Dylan: in other words, your second proof isn't really simpler than your first one. Right? IMO there should be no need to prove the statement by passing through $\Bbb E_1$-spaces---that's auxiliary. By the way, what is the first (historical) reference for the statement $\text{Map}_{\text{Mon}}(X,Y) \to \text{Map}_{E_1\text{-Mon}}(X,Y)$ is an equivalence when $X$ is cofibrant? You said that it is older than Lurie. $\endgroup$
    – John Klein
    Apr 15, 2013 at 0:40

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