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Let $X$ be a topological space and $A \subset X$ be a path connected subspace. Let $q: X \to X/A$ be the quotient map. Every example i work out it seems that the induced map $q_*: \pi_1(X) \to \pi_1(X/A)$ is a surjection. Is this always true? Are there some conditions under which it is true?

Thank you in advance!

kobe
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902
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  • Follow up question: If for example my space is super duper nice - like a simplicial complex, it is true correct? How about a CW complex? – 902 Sep 29 '16 at 07:29

2 Answers2

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It's not surjective in general. Consider the topologist sine curve $S$

$$S = \left\{\left(x, \sin \bigg(\frac{1}{x}\bigg)\right),\ 1/\pi\geq x>0\right\}$$

together with the interval $A = \{(0, t): |t|\leq 1\}$ and a curve $C$ joining this interval to the point $(1/\pi, 0)$. Call $X = A \cup S\cup C$. Then this is simply connected and $X/A$ is homeomorphic to $\mathbb S^1$, thus the induced map is zero.

Remark: To see that $X/A$ is homeomorphic to $\mathbb S^1$ Let $L = \{ (t,0) : t\in [0,1]\}$. Then $\mathbb S^1$ is homeomorphic to $B:=L\cup C$. Define $ F : X\to B$ by

$$F(x_1, x_2) = \begin{cases} (x_1,0) & \text{if } (x_1, x_2) \in A\cup S, \\ (x_1, x_2) & \text{if } (x_1, x_2) \in C. \end{cases}$$

One can check that $F$ is continuous and $f(A) = \{ (0,0)\}$. Thus $F$ descends to a continuous mapping $f: X/A \to \mathbb S^1$ which is bijective. Since $X/A$ is compact (Since $X$ is) and $\mathbb S^1$ is Hausdorff, $f$ is a homeomorphism.

  • Give me a moment to try and understand your example. – 902 Sep 29 '16 at 07:14
  • Isnt $\pi_1(X) = \mathbb{Z}$ in your example, and $\pi_1(X/A) = \mathbb{Z}$ also, or am i missing something? – 902 Sep 29 '16 at 07:20
  • $X$ is simply connected: No curve can pass through the topologist sine curve to $A$. @902 –  Sep 29 '16 at 07:23
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    oh, i think i understand now. Thanks so much! The reason why the problem does away when you quotient by $A$ is that we only need the path to pass the sine curve in the open interval and reach the destination at an infiniite time. – 902 Sep 29 '16 at 07:23
  • You are welcome. (And welcome to Mathstackexchange)@902 –  Sep 29 '16 at 07:25
  • Follow up question: If for example my space is super duper nice - like a simplicial complex, it is true correct? – 902 Sep 29 '16 at 07:26
  • How about a CW complex? – 902 Sep 29 '16 at 07:35
  • Sorry i still have one problem: Actually i realized my previous reasoning was not correct - Why is it true that $X/A$ is homeomorphic to $S^1$? Why is it's fundamental group not trivial for the same reason that $\pi_1X =0$? – 902 Sep 29 '16 at 08:05
  • I realized i never accepted your answer formally which I surely should have, and so, more than a full year after, I accept it! :) – 902 Feb 20 '18 at 04:57
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For CW complexes, you can do the following:

  1. Use the long exact sequence of homotopy groups to prove that $\pi_1(X)\rightarrow \pi_1(X,A)$ is surjective.
  2. Use excision for homotopy groups (Hatcher, Prop. 4.28) to prove that $\pi_1(X,A)\rightarrow \pi_1(X/A)$ is surjective.

EDIT: There is a shorter argument using Van Kampen's theorem: $X/A$ is homotopy equivalent to the union of $X$ and $CA$ (the cone of $A$) along $A$. As all three spaces are path connected, we have that the homomorphism $\pi_1(X)*\pi_1(CA)\to \pi_1(X/A)$ is surjective, and since $CA$ is contractible, we have that $\pi_1(X)\to \pi_1(X/A)$ is surjective as well. Notice that this works in more generality than CW-complexes.

user17786
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  • When you say that $X/A$ is homotopy equivalent to the union, what assupmtions on $A, X$ are made (if it is not a CW complex)? –  Oct 06 '16 at 11:14
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    That the inclusion $A\to X$ is a cofibration, i.e. $A$ is a neighbourhood deformation retractin $X$. – user17786 Oct 06 '16 at 11:22