# Fundamental group of a mapping torus

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?

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.

• 1. 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 $X$. Jun 5, 2023 at 11:02
• no, the codomain is just $X$. The domain is $X \sqcup X$, or $X \times \{0,1\}$. Jun 5, 2023 at 11:03
• Are you confusing domain and codomain? Jun 5, 2023 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? Jun 5, 2023 at 11:05
• yes (unless that point isn't in the image of $f$). Jun 5, 2023 at 11:06