Proof of the Group Completion Theorem [Hatcher], Fiberwise Suspension

Question

I'm reading Hatcher's proof of the group completion theorem, in A short exposition of Madsen-Weiss Theorem, Pages 36 ~ 42.

After Lemma D.3., Hatcher says that the total space is homotopy equivalent to $BM \vee BM$. I do not understand this.

As far as I understand, for a map $f : X \rightarrow Y$, the mapping cylinder is $M_f = X \times I \coprod Y \Big/ (x,1) \sim f(x)$ and from the way he defines fibrewise suspension, it looks like the fibrewise suspension is $M_f \Big/ (x, 0) \sim f(x)$ (So that we have $(x, 0) \sim (x, 1) \sim f(x)$). Is this correct?

So, if $f : X \rightarrow BM$ where $X$ is contractible, when we contract $X$ in $\Sigma_f X$, won't we get $BM \vee S^{1}$? [$X \simeq *$ is a point, so we get $S^{1}$ from $I$]

Also I do not understand why $\Sigma TM \rightarrow \Sigma_f ETM \rightarrow BM$ is a quasifibration. I don't see how he uses what he proved earlier (suspension of a homology isomorphism is a weak homotopy equivalence and Lemma D.1) [He says this just two sentences before Lemma D.3.]

$\textbf{Edit}$ : I believe he is doing the following :

He proves that the left action is a homotopy equivalence in the first paragraph on Page 40, i.e. $TM \rightarrow TM$ (multiplication by $m \in M$) is a weak homotopy equivalence. Taking suspension, $\Sigma TM \rightarrow \Sigma TM$ is a weak homotopy equivalence. In particular, right action by $m \in M$ is a homotopy equivalence, so, by Lemma D.1. (with $X = \Sigma TM$), $E\Sigma TM \rightarrow BM$ is a quasifibration.

Maybe $E \Sigma TM$ and $\Sigma_f ETM$ are somewhat related? But I can not see it.

Answer 1

As for the second question, let us try writing out explicitly the points of $\Sigma_f ETM$ and $E\Sigma TM$.

The first space is just $$I\times ETM/\sim$$ where $\sim$ identifies points with same image (on the $0$-layer and $1$-layer independently) so its points can be identified by a tuple $$(t, x)$$ with $x\in ETM$. Such $x$, by the definition of $ETM$ (which one can regard as the geometric realization of a certain category) is of the form $$(p, r, (m_1, \dots, m_k))$$ where $p$ is a point in the simplex $\Delta^k$ and $r$ is simply a label in the infinite telescope $TM$.

Similarly a point in $E\Sigma TM$ is a tuple $$(p, (t, r), (m_1, \dots, m_k))$$ where $p\in \Delta^k$ and $(t, r)\in \Sigma TM$ or equivalently $t\in I,\,\, r\in TM$.

The identifications between points in simplex of the space $E\Sigma TM$ comes from face and degeneracy maps ruled by the action of $M$ over $\Sigma TM$, which in turn is defined via the natural action of $M$ over the space $TM$ and extended to the suspension by not caring of the time coordinate (or we can say that $M$ is acting on $I$ trivially and taking the product action).

The action on $TM$ is precisely what rules the identification in $ETM$ and the free time coordinate is taken care of by the $\Sigma_f$ functor.

It seems to me that the map $$(\stackrel{\Sigma_f}{t}, \overbrace{(p, r, (m_1, \dots, m_k))}^{ETM}\,\,)\mapsto (p, \overbrace{(t, r)}^{\Sigma TM}, (m_1, \dots, m_k))$$ between the spaces $$I\times \left(\bigcup_{k} \Delta^k \times TM \times M^k\right) \rightarrow \bigcup \Delta^k \times \Sigma TM \times M^k$$ passes to the quotients $\Sigma_f ETM \rightarrow E\Sigma TM$ induced both by the relations of identifying simplexes and by the fiberwise suspension and gives an homeomorphism.

Answer 2

I can only answer the first question at this moment. If I understand Hatcher correctly, the fiberwise suspension is defined to be the double mapping cylinder, i.e. the pushout of $$X\times I\leftarrow X\amalg X\xrightarrow{f\amalg f} Y\amalg Y,$$ where the arrow to the left includes one summand into the $0$ coordinate and the other summand in the $1$ coordinate. Since double mapping cylinders preserve homotopy equivalences (see tom Dieck Proposition 4.2.1), if $X$ is contractible, $\Sigma_f X$ is homotopy equivalent to the pushout of $$I\leftarrow S^0\to Y\amalg Y,$$ which is homotopy equivalent to $Y\vee Y$.