New to algebraic topology.

Munkres (Topology, 2 ed.) in the last paragraph on page 332 says that "If $X$ is path-connected, all the groups $\pi_1(X,x)$ are isomorphic, so it is tempting to try to "identify" all these groups with one another and to speak of the fundamental group of the space X, without reference to base-point".

He goes on to say that there is no "natural way of identifying" these groups and "different paths...may give rise to different isomorphisms between these groups."

I just don't get this at all. If groups are isomorphic, they are the same algebraically. What does "different isomorphisms between groups" mean?

Again on the following page (334) he says that even in a path-connected space, the induced homomorphism map is an isomorphism, and "these groups are isomorphic, [but] they are still not the same group".

This is driving me nuts. The Klein-$V_4$ group is isomorphic to $\mathbb{Z_2} \times \mathbb{Z_2}$ and $D_4$; OK they do not "arise" naturally the same way, but they are the same algebraically and up to isomorphism.

My question is this: Agreed in a space not path-connected, you cannot dispense with the base-point; why is that a problem in a path-connected space?

Thank you in advance.

  • $\begingroup$ A group, by definition, is a set equipped with a certain binary operation. If the sets are different, groups can still be isomorphic. Now, read through the definitions of $\pi_1(X,x)$ and $\pi_1(X,y)$, $x\ne y$, and see if they are the same set or not. $\endgroup$ – Moishe Kohan Aug 18 '14 at 16:37
  • $\begingroup$ For example, the groups $\mathbb{Z}/3\mathbb{Z}=\{0,1,2\}$ and the abelian group $\{0,a,b\}$ with the sum given by $a+b=0, a+0=a, b+0=0$ are isomorphic, but there is no clear choice of what the isomorphism between these two groups should be. Should we map $1$ to $a$ or to $b$? $\endgroup$ – Dan Rust Aug 18 '14 at 16:39
  • $\begingroup$ If $x_0,x_1\in X$ and $\alpha:[0,1]\to X$ a path such that $\alpha(0)=x_0$ and $\alpha(1)=x_1$ then the map $\varphi_{\alpha}:\pi_1(X,x_0)\to \pi_1(X,x_1)$, $\varphi_{\alpha}([\gamma])=[\alpha\gamma\alpha^{-1}]$ is an isomorphism, but is not canonically, it depend of the choice of the path $\alpha$. $\endgroup$ – Hamou Aug 18 '14 at 16:47

Perhaps answering the first part of your question will shed light on your general discomfort here.

Just because two groups $G$ and $H$ are isomorphic does not mean that there is only one isomorphism between them. That is, there may be isomorphisms $\phi: G \to H$ and $\psi : G \to H$ with $\phi \neq \psi$. This is the point Munkres is making.

Given $x_0, x_1 \in X$, different homotopy classes of paths $x_0 \to x_1$ give rise to different isomorphisms $\pi_1(X, x_0) \to \pi_1(X, x_1)$. Thus, in order to find such an isomorphism, we must choose a class of paths from $x_0$ to $x_1$. This choice is why Munkres says there is no natural isomorphism between $\pi_1(X, x_0)$ and $\pi_1(X, x_1)$.

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  • $\begingroup$ I wish he would have clarified this with an example. What he is saying is that there may be different ways to define this isomorphism. And that I perfectly agree. Thanks a lot for clarifying. $\endgroup$ – Wulfgang Aug 18 '14 at 16:46
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    $\begingroup$ @Wulfgang What is also important to take away from this is that when you are given a path between $x_0$ and $x_1$ then there is a canonical isomorphism between the fundamental groups. $\endgroup$ – Dan Rust Aug 18 '14 at 16:48
  • $\begingroup$ @DanielRust not canonical, it depend of the choice of the path. $\endgroup$ – Hamou Aug 18 '14 at 16:54
  • $\begingroup$ @Hamou that's why I said when you have a path it is then canonical. This then relates quite closely to the fundamental groupoid of a space, where these possible isomorphisms induced by paths are the key objects of interest. $\endgroup$ – Dan Rust Aug 18 '14 at 17:00
  • $\begingroup$ This question confirms my view that a lack of a groupoidal approach to 1-dimensional homotopy theory easily leads to confusion. This groupoid approach has been advocated by me since 1968, see my web pages. $\endgroup$ – Ronnie Brown Aug 18 '14 at 21:21

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