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This is Problem 7-9 in Lee's Introduction to Topological Manifolds:

Suppose $X$ and $Y$ are connected topological spaces, and the fundamental group of $Y$ is abelian. Show that if $F,G: X \rightarrow Y$ are homotopic maps such that $F(x) = G(x)$ for some $x\in X$, then $F_*=G_* : \pi_1(X,x) \rightarrow \pi_1(Y,F(x))$. Give a counterexample to show that this might not be true if $\pi_1(Y)$ is not abelian.

I can't solve this problem.

  1. First of all, I wonder why connectedness condition is given not path connecteness condition. If $Y$ is not path connected, we can not say, I think, THE fundamental group of $Y$.

  2. Let $H:X \times I \rightarrow Y$ be a homotopy of $F$ and $G$. To prove $F_*=G_*$, put $[f]\in \pi_1(X,x)$ and we must show that $F \circ f \sim G \circ f$. $H(f(s),t)$ is not the desired one since the end points is not fixed. Putting $h(t) = H(f(0),t)$ and $k(t) = H(f(1),t)$, we can say $(F \circ f) \cdot k \sim h \cdot (G \circ f)$ by the square lemma. I can not progress at this point. Could I have any hint?

  3. If possible, I would like to have the counter example.

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  • $\begingroup$ Ad 2, note that $f(1) = f(0)$, so $k = h$. $\endgroup$ – Daniel Fischer Jul 9 '15 at 11:51
  • $\begingroup$ The word "connected" is a typo -- it should say "path connected." I've added this to my correction list. As for a hint on how to approach the problem, take a look at Lemma 7.45. $\endgroup$ – Jack Lee Jul 9 '15 at 15:57
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    $\begingroup$ Any space $Y$ with nonabelian fundamental group will serve as a counterexample. I leave it to you to pick the right $X$. In addition, you absolutely can say "the fundamental group of $Y$ based at $y$", whether or not $Y$ is path-connected. In fact this problem would remain completely correct if the word "connected" was dropped entirely. Anything relating to $\pi_1(Y,y)$ depends only on the path-component of $y$. $\endgroup$ – user98602 Jul 9 '15 at 17:41
  • $\begingroup$ Now I understand it. Thank you sir. $\endgroup$ – Jeong Jul 10 '15 at 11:20
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    $\begingroup$ @MikeMiller: You're right in a way, but without assuming path-connectedness, there might be an ambiguity about what's meant by saying "the fundamental group of $Y$ is abelian." It could be the case that the fundamental group of one path component of $Y$ is abelian, but $F(x)$ lies in another path component. That's why I included the path-connectedness hypothesis. $\endgroup$ – Jack Lee Jul 10 '15 at 16:35

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