Fundamental group of mapping cone of quotient map from suspension to reduced suspension

Question

Update: Thanks to Paul Frost, I realized there were mistakes in the question and my former proof. I moved the former proof to my answer below and modified the question.

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This is exercise 1.2.18(b) in page 55-56 of Hatcher's book Algebraic topology.

In this question: $X=\{0,1,\frac{1}{2}, \frac{1}{3},\cdots\}$, $SX$ is suspension and $\Sigma X$ is reduced suspension, where the suspension $SX$ is the quotient of $X\times I$ obtained by collapsing $X\times\{0\}$ to one point and $X\times\{1\}$ to another point and reduced suspension $\Sigma X$ is obtained from $SX$ by collapsing the segment $\color{blue}{\{0\}\times I}$.

Note: Reduced suspension depends on the choice of basepoint.

In this question, if we obtain reduced suspension from $SX$ by collapsing the segment $\{0\}\times I$, we get Hawaiian earring/The Shrinking Wedge of Circles in figure (4).

If we obtain reduced suspension from $SX$ by collapsing the segment $\{1\}\times I$, we get a space in figure (3) which is homotopy equivalent to $SX$.

Question:

Let $C$ be the mapping cone of the quotient map $SX→ΣX$.

Show that $π_1(C)$ is uncountable by constructing a homomorphism from $π_1(C)$ onto $∏_∞ \mathbb Z/\bigoplus_∞ \mathbb Z$.

Thanks for your time and effort.

Answer 1

You seem to claim that $SX$ is homotopy equivalent to the second space in your picture (which I shall denote by $S'X \subset \mathbb R^2$). This is not true. The yellow circle does not belong to $S'X$, thus $S'X$ is not compact. If you have any map $f : SX \to S'X$, then its image is compact and therefore must be contained in some $S'_n = \bigcup_{i=1}^n A_i$. This is a finite wedge of circles. We have $f = i_n f_n$ where $f_n : SX \to S'_n$ is the restriction of $f$ and $i_n : S'_n \to S'X$ denotes inclusion. If $g : S'X \to SX$ would be a homotopy inverse of $f$, then the identity on $\pi_1(SX)$ would factor through $\pi_1(S'_n)$ which is false.

However, there is no reason to replace $SX$ by another space. By the way, note that $\Sigma X$ is known as the Hawaiian earring. In Hatcher's Example 1.25 it is denoted as "The Shrinking Wedge of Circles".

As basepoint for $SX$ choose the midpoint $x_0$ of the black line segment and as basepoint for $\Sigma X$ choose the cluster point $y_0$ of the circles $B_i$. We have obvious pointed retractions $r_i : SX \to A_i$ (which project $A_j$ to the black line segment for $j \ne i$) and $s_i : \Sigma X \to B_i$ (which map $B_j$ to $y_0$ for $j \ne i$). This gives us group homomorphisms

$$\phi : \pi_1(SX,x_0) \to \prod_{i=1}^\infty \pi_1(A_i,x_0) = \prod_{i=1}^\infty \mathbb Z, \phi(a) = ((r_1)_*(a), (r_2)_*(a),\ldots),$$ $$\psi : \pi_1(\Sigma X,y_0) \to \prod_{i=1}^\infty \pi_1(B_i,y_0) = \prod_{i=1}^\infty \mathbb Z, \psi(b) = ((s_1)_*(b), (s_2)_*(b),\ldots) .$$ It is easy to see that $\psi$ is surjective, but $\phi$ is not. In fact, $\text{im}(\phi) = \bigoplus_{i=1}^\infty \mathbb Z$. This is true because all but finitely many $(r_i)_*(a)$ must be zero (otherwise the path representing $a$ would run infinitely many times through both endpoints of the black line segment, thus would have infinite length).

Obviously we have $\psi \circ q_* = \phi$, where $q : SX \to \Sigma X$ is the quotient map.

Now let us apply van Kampen's theorem. Write $C = U_1 \cup U_2$, where $U_1$ is obtained from $C$ by removing the tip of mapping cone and $U_2$ by removing the base $\Sigma X$. Both $U_k$ are open in $C$. We have

1. $U_1 \cap U_2 \approx SX \times (0,1) \simeq SX$ (thus $U_1 \cap U_2$ is path connected)

2. $U_1 \simeq \Sigma X$ (in fact, $\Sigma X$ is a strong defornation retract of $U_1$)

3. $U_2$ is contractible.

We conclude that $\Phi : \pi_1(U_1) * \pi_1(U_2) = \pi_1(\Sigma X) * 0 = \pi_1(\Sigma X) \to \pi_1(C)$ is surjective. Its kernel $N$ is the normal subgroup generated by the words of the form $(i_1)_*(c)(i_2)_*^{-1}(c)$, where $i_k : U_1 \cap U_2 \to U_k$ denotes inclusion and $c \in \pi_1(U_1 \cap U_2)$. Since $(i_2)_*^{-1}(c) = 0$, we see that $N$ is the normal closure of the image of the map $(i_1)_* : \pi_1(U_1 \cap U_2) \to \pi_1(U_1)$. But under the identifications $U_1 \cap U_2 \simeq SX$ and $U_1 \simeq \Sigma X$ we see that $(i_1)_*$ corresponds to $q_* : \pi_1(SX) \to \pi_1 (\Sigma X)$.

Hence $\pi_1(C) \approx \pi_1 (\Sigma X)/ N'$, where $N'$ is the normal closure of $\text{im}(q_*)$.

The surjective homomorphism $\psi' : \pi_1(\Sigma X) \stackrel{\psi}{\rightarrow} \prod_{i=1}^\infty \mathbb Z \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$ has the property $\psi' \circ q_* = 0$, thus $\text{im}(q_*) \subset \ker(\psi')$. Since $\ker(\psi')$ is a normal subgroup, we have $N' \subset \ker(\psi')$, thus $\psi'$ induces a surjective homomorphism $\pi_1(C) \approx \pi_1 (\Sigma X)/ N' \to \prod_{i=1}^\infty \mathbb Z / \bigoplus_{i=1}^\infty \mathbb Z$.

Answer 2

Your construction of $\varphi$ looks right to me (in the sense that I think it could appear in a full proof). I don’t think you have sufficiently proven that it is well-defined. I think this is the tricky part you felt you were missing.

I think you could perhaps add a few more words as to why the paths you use to show $\varphi$ is surjective are not contractible.

Edit:

The question changed. I feel like it’s a bit confused and it’s not really at all clear to me that $\varphi$ is well-defined. I feel like there’s too much if $x$ then $y$ otherwise $z$ (especially as sometimes this is used to split up cases and sometimes for an argument by contradiction). Perhaps a better way to structure this would be as follows:

1. Define the function $\psi$ mapping $\pi_1(\Sigma X)$ to $\prod_\infty \Bbb Z$
2. Show that it is a well-defined surjective homomorphism
3. Let $r : \prod_\infty\Bbb Z\to \prod_\infty\Bbb Z/\bigoplus_\infty\Bbb Z$ be the quotient map
4. Prove that $r\psi$ may be extended to a well-defined homomorphism $\phi : \pi_1(C_q)\to \prod_\infty\Bbb Z/\bigoplus_\infty\Bbb Z$.
5. You get surjectivity from step 2

But actually I feel like you should be able to use some theorem for this instead. The theorem I am imagining is one which would say something like $\pi_1(C_q) = \pi_1(\Sigma X)/\pi_1(SX)$ and then you need the theorem from groups that a homomorphism $f: G \to H$ induces a homomorphism $G/N \to G/f(N)$. Maybe you don’t have that theorem but do you have Seifert-van Kampen yet? Maybe you could use that to your advantage here.

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Edit 2:

I looked up the exercise in Hatcher. The exercise is in the first block after the proof and statement of van Kampen’s theorem (which I have perhaps confusingly referred to as Seifert-van Kampen). I think the exercise wants you to use the theorem. You should take one set to be $\Sigma X$ plus the mapping cylinder of $q,$ and the other to be $CSX$, so the intersection is $S X$. My guess is that the point of the exercise is practicing van Kampen’s theorem (and I guess a bit about how it relates to mapping cones and these spaces).

Answer 3

Alternative method:

Let $U=C(SX)$, $V$ be mapping cylinder of $q$, $V=SX \times I \sqcup \Sigma X /\sim$, i.e. $(x,1)\sim q(x)$ for $x\in SX$.

Let $W$ to be space by gluing $U$ and $V$ along $SX$.

$V$ deformation retracts to $\Sigma X$, so $C_q=\Sigma X \cup_q C(SX) \simeq W$.

$\pi_1(U)=0$. Let $i_1: SX \hookrightarrow C(SX)=U$, $i_2: SX \hookrightarrow V$.

$i_{1*}=0$, $i_{2*}:\pi_1(SX)\to\pi_1(V)\cong\pi_1(\Sigma X)$.

From van Kampen's theorem, $\pi_1(W)\cong\pi_1(U)*\pi_1(V)/N$, $N$ is generated by $i_{1*}(w)i_{2*}(w^{-1})$ for all $w\in \pi_1(SX)$. $\pi_1(W)\cong \pi_1(\Sigma X)/\pi_1(SX)$.

Let $\rho$ be the surjective homomorphism $\pi_1(\Sigma X) \to \prod_\infty \mathbb Z$ given in page 49 of Hatcher.

$\color{red}{\text{$\rho$ maps $\pi_1(SX)$ to $\bigoplus_\infty \mathbb Z$} \ (*) }$ , so $\rho$ induces a surjective homomorphism

$\pi_1(\Sigma X)/\pi_1(SX) \to \prod_\infty \mathbb Z/\bigoplus_\infty \mathbb Z$. $\quad\Box$

Claim $(*)$ in red is remained to check.

I tried to give an explicit expression of $\rho$ to prove claim $(*)$ in the question, but I couldn't fully prove the well-defineness of the $\rho$ I constructed.

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My former proof (wrong):

$1$. Yellow circle doesn't belong to $SX$. It means the lower bound of circles in $SX$.

Circles in $\Sigma X$ have no such lower bound. Region containing "$\cdots$" means there're countably many circles in it.

From outside to inside, circles in $SX$ and $\Sigma X$ are denoted by $A_n$ and $B_n$ with common point $x_0$ and $y_0$.

Quotient map is $q:SX \to \Sigma X$, mapping cone $C_q=\Sigma X \cup_q CSX$.

$2$. Choose basepoint of loops at $x_0\sim y_0$. $A_i\sim B_i$, so loop around $A_i$ can be considered as loop around $B_i$.

For $[f]\in \pi_1(C_q)$, if $f$ wraps $a_n$ times around circle $B_n$ clockwise, let $\tilde f: I \to SX$ wrap $a_n$ times around circle $A_n$ clockwise the same ways as $f$. Note that $q \circ\tilde f=f$.

Define $\varphi: \pi_1(C_q)\to \prod_\infty \mathbb Z/\bigoplus_\infty \mathbb Z,\ [f]\mapsto \overline{(a_n)_{n=1}^{\infty}}$.

$3$. If $(a_n)_{n=1}^{\infty} \in \bigoplus_\infty \mathbb Z$, then only finite $a_n$ is nonzero and $\tilde f$ is continuous on $I$, so $\tilde f$ is indeed a loop in $C_q$.

$[\tilde f]=[q\circ \tilde f]=[f]$. Note that $A_n\subset SX \subset CSX$ is contractible, so $[\tilde f]=0=[f]$, $f$ is nullhomotopic. $\pi_1([f])=\overline 0$.

This is consistent with $\overline{(a_n)_{n=1}^{\infty}}=\overline 0$ in $\prod_\infty \mathbb Z/\bigoplus_\infty \mathbb Z$.

If $(a_n)_{n=1}^{\infty} \in \prod_\infty \mathbb Z - \bigoplus_\infty \mathbb Z$, then $\tilde f$ is not continuous at $1$, so it's not a loop in $SX$ and $\color{red}{\text{ $f$ isn't nullhomotopic}}$.

$3$. $\varphi: \pi_1(C_q)\to \prod_\infty \mathbb Z/\bigoplus_\infty \mathbb Z$ is well-defined.

If $[f]=[g]$ in $\pi_1(C_q)$, $f, g$ wraps $a_n, b_n$ times around circle $B_n$ clockwise, then $f\circ g^{-1}$ wraps $a_n - b_n$ times around circle $B_n$ clockwise.

$f\circ g$ is nullhomotopic, so $(a_n-b_n)_{n=1}^\infty\in \bigoplus_\infty \mathbb Z$.

$\varphi([f])=\overline{(a_n)_{n=1}^{\infty}}=\overline{(a_n-b_n+b_n)_{n=1}^{\infty}}=\overline{(a_n-b_n)_{n=1}^{\infty}}+\overline{(b_n)_{n=1}^{\infty}}=\overline{(b_n)_{n=1}^{\infty}}=\varphi([g])$.

$\varphi$ is a well-defined surjective homomorphism. $\quad\Box$