3
$\begingroup$

Consider the category $\mathrm{Mon}(\mathbf{Top})$ of topological monoids, together with the model structure transferred along the adjunction $F:\mathbf{Top}\rightleftarrows \mathrm{Mon}(\mathbf{Top}):U$, where we consider the Quillen model structure on spaces. Moreover, let $\times$ be the direct product of topological monoids.

Let $\imath\colon M\to M'$ be a cofibration of topological monoids and let $N$ be another topological monoid (cofibrant, if necessary). Is it true that the map $\imath\times\mathrm{id}_N\colon M\times N\to M'\times N$ is still a cofibration?

(Originally, I thought that $\mathrm{Mon}(\mathbf{Top})$ even satisfies the pushout-product axiom with respect to $(\times,1)$, but this is maybe too much to ask for?)

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ I don't think so. The inclusion $1\to N$ of the trivial subgroup into a free monoid on one generator is a cofibration. Let $G$ be a group of order two. I don't think that the inclusion $G=1\times G\into N\times G$ is a cofibration. For a monoid $P$, extending a map $G\to P$ to a map $N\times G\to P$ means choosing an element $y\in P$ that commutes with a given element $x\in P$ whose square is $1$. But homotopies cannot always be lifted: there can be a path $x_t$ with $x_t^2=1$ and an element $y_0$ commuting with $x_0$ such that for $t>0$ there is nothing near $y_0$ commuting with $x_t$. $\endgroup$
    – Tom Goodwillie
    Nov 22, 2020 at 22:24
  • $\begingroup$ Thank you! Unfortunately, I don’t have a good intuition for acyclic fibrations in $\mathrm{Mon}(\mathbf{Top})$, but I assume that $\varepsilon_0\colon M^I\to M$ is always an acyclic fibration, where $M^I$ is given the pointwise multiplication? And secondly, is it true that $G$ is cofibrant? Clearly, $1\to G$ left lifts against all $M^I\to M$, but is this enough? $\endgroup$
    – FKranhold
    Nov 23, 2020 at 9:46

1 Answer 1

Reset to default
1
$\begingroup$

I think I found a counterexample of which I am sure that the monoids we started with are as nice as we could wish for: Start with the cofibration $\imath\colon 1\to \mathbb{N}$. We will show that $\mathrm{id}_{\mathbb{N}}\times\imath\colon \mathbb{N}\to \mathbb{N}\times \mathbb{N}, p\mapsto (p,0)$ is not a cofibration.

Consider the topological group $G:=\mathbb{R}^3$ with $(a,b,c)\cdot (a',b',c')=(a+a',b+b',c+c'+ab')$. Then the free path space $G^I$ is again a topological group by pointwise multiplication, and the evaluation $\varepsilon\colon G^I\to G, \alpha\mapsto \alpha(0)$ is an acyclic fibration since its underlying map of spaces is one.

Now consider $f\colon \mathbb{N}\times\mathbb{N}\to G,(p,q)\mapsto (0,q,0)$ which is clearly a homomorphism of topological monoids, and consider $h:\mathbb{N}\to G^I$ defined by $h(p)(t)=(tp,0,0)$. Note that the corresponding square commutes, i.e. we have $(\varepsilon\circ h)(p) = (0,0,0)=f(p,0)=(f\circ (\mathrm{id}\times\imath))(p)$.

However, there is no lift: Note that a homomorphism $H\colon \mathbb{N}\times\mathbb{N}\to G^I$ is the same as a choice of paths $\alpha=H(1,0)$ and $\beta=H(0,1)$ such that $\alpha\cdot\beta=\beta\cdot\alpha$ holds pointwise. The commutativity of the two triangles then implies that $\alpha=h(1)$ and $\beta(0)=(0,1,0)$. However, continuity of $\beta$ implies that there is a $t>0$ and a $s>0$ as well as $x,y\in\mathbb{R}$ such that $\beta(t)=(x,s,y)$, which gives $$\alpha(t)\cdot\beta(t)=(t,0,0)\cdot (x,s,y)\ne (x,s,y)\cdot (t,0,0)=\beta(t)\cdot \alpha(t).$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.