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Let $(X,\mu,e)$ be an $H$-space (so $\mu:X\times X\to X$ is a continuous map and $e\in X$ is a base point such that $\mu(\cdot,e)\simeq id_X\simeq\mu(e,\cdot)$) and a CW-complex such that $e$ is a $0$-cell for the CW-structure on $X$. I am trying to show that there exists a continuous map $\tilde\mu:X\times X\to X$ such that $\tilde\mu(x,e)=x=\tilde\mu(e,x)$. Supposedly it makes sense to use the homotopy extension property for this, but I fail to see how: I would think that starting at a and building up $X$ from there, you can use the HEP to build the map $\tilde\mu$. Besides the fact that I don't see why that map would satisfy the condition that $\tilde\mu$ should satisfy, and you only have one 0-cell, so how can you get the whole CW-complex starting from this?

Any help/hints are appreciated.

Edit: the answer below seems to be complete, but I can't quite digest it because I don't know what this folding map is.

Edit2: Apparently, the answer below seems to consider $X\bigvee X$ as the subspace $X\times\{e\}\sqcup\{e\}\times X$ of $X^2$, so that the folding map is just $(x,e)\mapsto x$, $(e,x) \mapsto x$. I don't see how this answer uses the fact that $e$ is a 0-cell.

B. Pasternak
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1 Answers1

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Let $X$ be a CW complex and $e$ a zero cell of $X$.

Lemma 1 The inclusion $\{e\}\hookrightarrow X$ is a closed cofibration.

Proof It's true more generally that the inclusion of any subcomplex of a CW complex is a closed cofibration. $\quad\blacksquare$

Write $$X\vee X=X\times\{e\}\cup \{e\}\times X.$$ The inclusion $j:X\vee X\rightarrow X\times X$ is a closed embedding.

Lemma 2 The embedding $j$ is a closed cofibration.

Proof It is a standard fact that the product of two closed cofibrations is again a closed cofibration. $\quad\blacksquare$

Lemma 3 The inclusion $\{e\}\times\{e\}\hookrightarrow X\times X$ is a closed cofibration.

Proof Since $X$ is a CW complex, $X^2$ is perfectly normal and locally contractible. Thus the methods here can be applied. $\quad\blacksquare$

With the preliminaries out of the way let us begin. We are given a map $\mu:X\times X\rightarrow X$ and homotopies $F^l_t:\mu(-,e)\simeq id_X$ and $F^r_t:\mu(e,-)\simeq id_X$.

Lemma We may assume that $\mu(e,e)=e$.

Proof If this is not already true then choose a path $\lambda:I\rightarrow X$ from $\lambda(0)=\mu(e,e)$ and $\lambda(1)=e$. Such a path always exists (for instance, one may be extracted from either of the homotopies $F^l_t,F^r_t$). Viewing $\lambda$ as a homotopy apply Lemma 3 to find a homotopy $G:\mu\simeq\mu'$ where $\mu'(e,e)=e$. We obtain homotopies ${F^l_t}':\mu'(-,e)\simeq id_X$ and ${F^l_t}':\mu'(e,-)\simeq id_X$ by forming composite homotopies of $F^l_t,F^r_t$ with $G$. $\qquad\blacksquare$

Lemma We may assume that $F^l_t(e)=e$ and $F^r_t(e)=e$ for all $t\in I$.

Proof The proof of this is quite long. It is a special case of the proposition on pg.46 of May's book A Concise Course in Algebraic Topology. $\quad\blacksquare$

At this stage we have replaced $\mu$ with a pointed map and the homotopies $F^l_t,F^r_t$ with pointed homotopies. Because they are pointed, the homotopies assemble to give a homotopy $$F:\mu\circ j\simeq \nabla:X\vee X\rightarrow X$$ which restricts to $F^l$, resp. $F^r$ on the left, resp. right wedge summands.

Now use Lemma 2 to apply the HEP for $j$ to the homotopy $F$. This gives a homotopy $G:X\times X\times I\rightarrow X$ starting at $\mu$ and ending at some map $\tilde\mu:X\times X\rightarrow X$ which satisfies $\tilde\mu\circ j=\nabla$ strictly. That is, $\tilde\mu(x,e)=x=\tilde\mu(e,x)$ for each $x\in X$.

Tyrone
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  • How is the folding map defined? – B. Pasternak Dec 13 '17 at 09:10
  • The wedge product $X\vee Y$ of two based spaces $X,Y$ is the categorical coproduct in the category of based spaces (although you can think of it as the subset $X\times \cup \times Y\subseteq X\times Y$). This means that a map $f:X\vee Y\rightarrow Z$ is completely determined by a pair of maps $f_X:X\rightarrow Z$ and $f_Y:Y\rightarrow Z$. We often write $f=(f_X,f_Y)$. When $X=Y=Z$ the fold map is defined by $\nabla=(id_X,id_X)$. Explicitly $\nabla(x,)=x=\nabla(,x)$. – Tyrone Dec 13 '17 at 10:48
  • I see. But why does $(X\times X,X\bigvee X)$ have the HEP? – B. Pasternak Dec 13 '17 at 11:36
  • Since CW complexes are paracompact Hausdorff the inclusion of a 0-cell basepoint is a closed cofibration $e\hookrightarrow X$. As is the identity $X\xrightarrow{=}X$. The product of cofibrations is again a cofibration, as is the union of cofibrations. Hence $X\vee X=X\times \ast\cup\ast\times X\hookrightarrow X\times X$ is a closed cofibration. – Tyrone Dec 13 '17 at 13:13
  • The issue is the closed cofibration ↪ CW is sufficient but not necessary – Jim Stasheff Oct 13 '20 at 20:48
  • Why the two homotopies $F_t^l$ and $F_t^r$ assemble to a homotopy? This means the basepoint $e$ is preserved through $F_t^l$ and $F_t^r$, i.e., $X$ is an H-space. – Ted Mar 08 '25 at 12:33
  • You start out by asserting that $\mu$ extends $\nabla$ over $j$. This is not given; in fact it's what you're trying to show, since it precisely amounts to saying that $\mu(e, x) = \nabla((e, x)) = x = \nabla((x, e)) = \mu(x, e)$! As written, your answer asserts the sequent in order to prove it. – Ben Steffan Mar 08 '25 at 14:21
  • @BenSteffan my assumption is supposed to be that $\mu$ extends $\nabla$ up to homotopy, which is what the rest of the answer is based on. I'll clarify it when I get a chance. – Tyrone Mar 08 '25 at 20:31
  • @Tyrone I think the proof is correct, but there are redundant parts. It suffices to prove 1) we can assume $\mu(e,e)=e$ and 2) apply May's proposition to the diagram $e\xrightarrow{i}X\xrightarrow{\mu(\cdot,e)}X\xleftarrow{j}e$. – Ted Mar 09 '25 at 08:39