Zeroth homotopy group: what exactly is it?

Question

What are the elements in the zeroth homotopy group? Also, why does $\pi_0(X)=0$ imply that the space is path-connected?

Thanks for the help. I find that zeroth homotopy groups are rarely discussed in literature, hence having some trouble understanding it. I do understand that the elements in $\pi_1(X)$ are loops (homotopy classes of loops), trying to see the relation to $\pi_0$.

One can simply define $\pi_0$ of a space to be the set of path components of the space. The answers you are getting are using general definitions that work for all degrees (and which, i guess, you have not yet beeen introduced to) and specializing it to zero, but if you do not need (or know!) that generality, you can go quite a long way using the definittion I mentioned (which is, of course, equivalent to the others) — Mariano Suárez-Álvarez, Jan 02 '16 at 06:14
@MarianoSuárez-Alvarez thanks for your answer! Why is $\pi_0$ not a group? — yoyostein, Jan 02 '16 at 06:30
@Mauro I just read up: it is the set of two points at the end of a line segment — yoyostein, Jan 02 '16 at 06:32
@yoyostein, well, because it isn't! :-) The fact that $\pi_1$ is a group is what should surprise! — Mariano Suárez-Álvarez, Jan 02 '16 at 06:38

score 20 · Accepted Answer · answered Jan 02 '16 at 06:14

This is the definition.

So $\pi_0$ is the homotopy classes of maps from two points ($S^0$) to $X$, where the first point is mapped to the base point. Clearly only the path connected component matters for the second point (since a path connecting the two points defines a homotopy between two such maps).

$\pi_0$ being trivial implies that there is a path between any point and the base point, i.e. $X$ is path-connected.

Anubhav Mukherjee · Answer 2 · 2016-01-02T06:16:40.310

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For a space $X$ and a base point $x_0$ , define $\pi_n(X,x_0)$ to be the homotopy class of maps $f:(I^n, \partial I^n) \to (X,x_0)$ , where homotopies $f_t$ are required to satisfy $f_t(\partial I^n)= x_0$ . this definition can be extended to the case n=0 by taking $I^0$ to be a point and its boundary to be an empty set, so $\pi_0(X,x_0)$ is just the set of path connected component.

edited Jan 02 '16 at 06:16

answered Jan 02 '16 at 06:10

Anubhav Mukherjee

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score 3 · Answer 3 · edited Apr 24 '17 at 18:54

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Just a slight rephrase: you can consider $\pi_0(X)$ as the quotient set of the set of all points in $X$ where you mod out by the equivalence relation that identifies two points if there is a path between them.

edited Apr 24 '17 at 18:54

Ramil

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answered Apr 24 '17 at 18:07

user440042

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Zeroth homotopy group: what exactly is it?

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