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.
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 $\pi_1(X, x_1)$.