2

I've read through similar questions and have some additional questions.

Exercise: Let $f:X\rightarrow X$ be a self-mapping. Let $X$ be simply connected. Define mapping torus $T_f$ as pushout of

$$\require{AMScd} \begin{CD} X\times \{0,1\} @>{i}>> X \times [0,1]\\ @V{\text{id}_x \coprod f}VV @VV{}V\\ X @>>{}> T_f\end{CD}$$

with $i$ inclusion map and $\text{id}_x \coprod f: (x,0) \mapsto x, (x,1) \mapsto f(x)$. Calculate $\pi_1(T_f)$.

Question 1: How does the map $\text{id}_x \coprod f$ has $X$ as its codomain? Shouldnt it be two copies of $X$, since we literally map $X\times \{0\}$ to $X$ and $X\times \{1\}$ to $f(X)$?

Question 2: Are simply connected spaces homeomorphic to $D^n$ or $S^n$? In other words, can I assume that $f$ has a fixed point?

Question 3: I want to use van Kampen theorem to calculate $\pi_1(T_f)$. What is the best way to decompose $T_f$ in two connected subspaces? I wanted to choose $(x,\frac{1}{2})$ as my basepoint and $U=X\times [0,\frac{2}{3})$, $V=X\times (\frac{1}{3},1)$.

But this is where I lose the grip on the reality because for me, $\pi_1(U)$ and $\pi_1(V)$ should be equal to $\{1\}$ since they're homotopy equivalent to $X$ and I don't undestand what's wrong with my reasoning...

Another thing I'm not sure of is whether I can include $X\times \{1\}$ in $V$ or not because if I do then the intersection $U\cap V$ wouldn't be path-connected, right?

Question 4: In this question, the author decomposes $T_f$ as $U=X\times( 0,1)$ and $V= (X \times [0,1/3)) \cup (X \times (2/3,1]) \cup (N \times I)$. What does $(N\times I)$ mean? Why don't $V= (X \times [0,1/3)) \cup (X \times (2/3,1])$ work?

average math enjoyer
  • 150

1 Answers1

0

  1. The domain of the left hand map is what you say it should be, so what's the question there?

  2. No, there are many different simply connected spaces that are not homeomorphic to a disk.

  3. You're forgetting that the 'ends' of the cylinder are glued together. This is a mapping torus, not a mapping cylinder. The details for the decomposition to use for van Kampen are provided in this question.

Dan Rust
  • 30,108
  • If I were to be precise I would identify the codomain as $X\coprod X$ where $X\times {0}$ maps to the first $X$ and $X\times {1}$ maps to the second $X$.
    • – average math enjoyer Jun 05 '23 at 11:02
  • no, the codomain is just $X$. The domain is $X \sqcup X$, or $X \times {0,1}$. – Dan Rust Jun 05 '23 at 11:03
  • Are you confusing domain and codomain? – Dan Rust Jun 05 '23 at 11:05
  • So they really do map to the same $X$? I guess each point in the codomain has two points in the domain? – average math enjoyer Jun 05 '23 at 11:05
  • yes (unless that point isn't in the image of $f$). – Dan Rust Jun 05 '23 at 11:06
  • Maybe, that's what I'm trying to find out. Then the map is surjective but not injective? – average math enjoyer Jun 05 '23 at 11:06
  • At least one point has two preimages, so it's never injective. It's always surjective because every point $x$ has $(x,0)$ as a preimage. – Dan Rust Jun 05 '23 at 11:08
  • But if they overlap on two disjoint open intervals, then the intersection isn't path connected anymore and van Kampen fails?
    • – average math enjoyer Jun 05 '23 at 11:12
  • Apologies, I was thinking of Mayer-Vietoris. – Dan Rust Jun 05 '23 at 11:23
  • edited my answer – Dan Rust Jun 05 '23 at 11:32
  • I've read the answer you linked and added one more question based on it to my post. – average math enjoyer Jun 05 '23 at 12:09
  • $N$ is a contractible neighbourhood of the base point, as mentioned in the first paragraph of the question. – Dan Rust Jun 05 '23 at 13:00