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I am having trouble with Hatcher's Algebraic Topology P39, Problem 18:

Show that if a space $X$ is obtained from a path-connected subspace $A$ by attaching a cell $e^n$ with $n ≥ 2$, then the inclusion $A \rightarrow X$ induces a surjection on $π_1$.

The given hint is to follow the proof of Proposition 1.14, which states that $\pi_1(S^n) = 0$ if $n \geq 2$. The proof of 1.14 roughly goes: any given loop can be moved to avoid any specific point. Then, if a loop is disjoint from a point $x$, we can deformation retract $S^n - x$ onto a single point, thus killing the loop.

Applying this method to Problem 18 seems easy: I can show that if a loop is disjoint from any point $y \in e^n \backslash \delta e^n$, then I can deformation retract $e^n \backslash y$ onto $\delta e^n \subseteq A$. And, moving the loop off of $y$ is doable. Thus, any loop in $X$ becomes a loop in $A$. However, this doesn't get me to the result; loops in $X$ can still travel around $\delta e^n$, which adds connections between points in $A$ that didn't exist before.

To make this visual, I have a "counterexample" to the problem. 18a says "The wedge sum $S^1 ∨ S^2$ has fundamental group $\mathbb{Z}$." However, using the result from the problem, I can construct:

Let $A$ consist of two points $a$ and $b$, and a line between them. Then, attach $D^2$ by sending $e^{i \pi k} \rightarrow a$ for $0 \leq k < 1$, and $e^{i \pi j} \rightarrow b$ for $1 \leq j < 2$. This turns $D^2$ into a sphere, with $a$ and $b$ being two points infinitesimally close to each other on its surface, so our construction is homotopy equivalent to $S^1 ∨ S^2$. However, $\pi_1(A) = 0$, while $\pi_1(X) = \mathbb{Z}$.

So my question(s) is:

  1. How to actually prove Problem 18?
  2. What is wrong with my counterexample?
  3. Are $a$ and $b$ actually sent to infinitesimally close points?
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  • $\begingroup$ What do you possibly mean by "infinitesimally close points"? $\endgroup$ Commented Dec 2, 2013 at 4:54
  • $\begingroup$ I was imagining it like a sphere, with $a$ a fixed point and $b$ as a point distinct from $a$, but such that any open set containing $a$ would have to contain $b$ also. I am not entirely sure this is meaningful. $\endgroup$
    – Kevin Yin
    Commented Dec 2, 2013 at 4:59
  • $\begingroup$ I suggest that you review the definition of Hausdorff spaces and the basic examples (like the sphere!) before continuing. This is treated in any general topology textbook. You will save yourself lots of pain. $\endgroup$ Commented Dec 2, 2013 at 5:07
  • $\begingroup$ I am aware that the sphere is Hausdorff, and that having $a$ and $b$ without disjoint neighborhoods is impossible if they exist on the sphere. However, I am having trouble visualizing what the attached space is supposed to be, and whether $a$ and $b$ end up as distinct points or not, so I do not see where the error is, although I am sure it exists. $\endgroup$
    – Kevin Yin
    Commented Dec 2, 2013 at 5:13
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    $\begingroup$ The second paragraph of Chapter 0 of the book says: «To avoid overusing the word ‘continuous’ we adopt the convention that maps between spaces are always assumed to be continuous unless otherwise stated.» $\endgroup$ Commented Dec 2, 2013 at 5:33

1 Answer 1

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Reading the comments it seems your second and third questions have been answered. I address the first question.

The hint suggests using Lemma 1.15, which reads:

If a space $A$ is the union of a collection of path-connected open sets $A_\alpha$, each containing the basepoint $x_0\in A$ and if each intersection $A_\alpha\cap A_\beta$ is path-connected, then every loop in $A$ at $x_0$ is homotopic to a product of loops each of which is contained in a single $A_\alpha$.

Letting $A_\alpha=e^n-x$ where $x\in e^n-\partial e^n$, and $A_\beta=X$, we see that $A_\alpha\cap A_\beta=\partial e^n$ which is path connected and $A=A_\alpha\cup A_\beta$: the lemma will apply. Every loop in $A$ at $x_0$ is some product of loops $p_1\cdots p_n$, which is homotopic to a product of loops in $X$; since any $p_i\in\pi_1(A_\alpha)$ is nullhomotopic ($A_\alpha$ deformation retracts onto $S^k$, $k\ge 2$, then use Proposition 1.14). This is exactly what was required.


A similar result

After Section 1.2, we can use the van Kampen theorem to get a similar result: a CW complex $X$ and its two-skeleton $X^2$ have isomporphic fundamental groups.

Proof. Since $X^n$ is the quotient space of the disjoint union $X^{n-1}\coprod_\alpha D_\alpha^n$ with identifications $x\sim\phi_\alpha(x)$ for $x\in\partial D_\alpha^n$, it suffices to show that $\pi_1(X^{n-1}\coprod D^n)\cong\pi_1(X^{n-1})$ for $n\ge 3$.

Let $x\in D^n-\partial D^n$, and $A_\alpha=D^n-x$; and $A_\beta=X^{n-1}$. Then $X^{n-1}\coprod D^n=A_\alpha\cup A_\beta$, and $A_\alpha\cap A_\beta=\partial D^n$ which is isomorphic to $S^{n-1}$ which is path connected. By the van Kampen theorem, $$\pi_1(X^n)\cong \pi_1(A_\alpha)\ast\pi_1(A_\beta)\cong\pi_1(S^{n-1})\ast\pi_1(X^{n-1})\cong \pi_1(X^2),$$ using an induction argument, and noting that $\pi_1(S^{n-1})\cong 0$ for $n\ge 2$. In particular, since $\pi_1(X^n)\cong\pi_1(X^2)$ for all $n$, $\pi_1(X)\cong\pi_1(X^2)$.

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  • $\begingroup$ I have several questions about this solution. First off, $A$ is a subspace of $X$ in the problem so I think all of the $A$'s and $X$'s should be switched in the solution. Now it is not true that $X=A_\alpha\cup A_\beta$ because it does not contain $x$. Also, how can we apply the Lemma is $A$ if not open? $\endgroup$
    – 1234
    Commented Jul 2, 2016 at 23:55
  • $\begingroup$ the indices seem to be off as well: shouldn't $A_\alpha = e^n-x$ deform retract to the $(n-1)$-sphere instead? $\endgroup$ Commented Jun 24, 2023 at 12:53

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