# Are cofibrations in topological monoids preserved by forming the product with the identity?

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?)

• 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 $x_t$. Nov 22 '20 at 22:24
• 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? Nov 23 '20 at 9:46

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 $$\alpha(t)\cdot\beta(t)=(t,0,0)\cdot (x,s,y)\ne (x,s,y)\cdot (t,0,0)=\beta(t)\cdot \alpha(t).$$