# What's the meaning of $\pi_0$ in Algebraic Topology? [duplicate]

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In my general topology class, we introduced some basic algebraic topology. My professor mentioned the term $\pi_0$ together with path components of $X$, but I forgot what he said. Does anyone know what is it? And he also mentioned $\pi_{\leq 1}$ and $\pi_2$. What are these things? We use Munkres' Topology. In section 52, he mentions "There are indeed groups $\pi_n (X,x_0)$ for all $n \in \Bbb Z_+$, but we shall not study them in this book. They are part of the general subject called homotopy theory."

## marked as duplicate by Ethan Bolker, GNUSupporter 8964民主女神 地下教會, Aloizio Macedo♦, Lee Mosher, ahulpkeApr 28 '18 at 1:02

• $\pi_0(X)$ is the set of path-components of $X$. $\pi_1$ and $\pi_2$ are homotopy groups; $\pi_1$ is the fundamental group. – Lord Shark the Unknown Apr 27 '18 at 16:25

$\pi_0(X)$ is the set of all path components of $X$, i.e. we introduce the equivalence relation on $X$: $x\sim y$ if and only if there exists a (continuous) path $\lambda:[0,1]\to X$ such that $\lambda(0)=x$ and $\lambda(1)=y$. Then

$$\pi_0(X):=X/\sim$$

Now let $S^n\subseteq\mathbb{R}^{n+1}$ ne the $n$-dimensional standard sphere and take a fixed point on that sphere, say $P=(1,0,\ldots, 0)$. If $X$ is a topological space and $x_0\in X$ then define

$$L_n(X,x_0):=\{f:S^n\to X\ |\ f\text{ is continuous and }f(P)=x_0\}$$

On that set define the relation:

$$f\sim g\text{ if and only if }f\text{ is homotopic to }g$$

Then by definition

$$\pi_n(X,x_0):=L_n(X,x_0)/\sim$$

For any $n>0$ the set $\pi_n(X,x_0)$ can be turned into a group widely known as the $n$-th homotopy group. For $n=1$ the group $\pi_1(X,x_0)$ is also know as the fundamental group of $(X,x_0)$. I'm not gonna go into details how the group structure is defined and why the structure is intersting, you can read more about it here.

The last important fact I'm going to mention is that if $X$ is path connected and $x_0,x_1\in X$ then $\pi_n(X,x_0)\simeq\pi_n(X,x_1)$ as groups. In that case we often write $\pi_n(X)$ for simplicity.

Generally speaking, $$\pi_n(X, x_0)$$ captures the homotopy information of continuos maps $$f:S^n \to X$$, where $$f(\star)=x_0$$ and $$\star$$ is a distinguished point of $$S^n$$ (in the case $$n=1$$, that's usually the point that corresponds to the identification $$1\sim 0$$ on the interval $$[0,1]$$, and you can make similar constructions for higher dimensions).

Thus the case $$n=0$$ tells you about homotopy of maps $$\{\star, 1\}\to X$$ sending $$\star$$ to $$x_0$$ and $$1$$ somewhere else. There are, up to homotopy, as many choices of where to send $$1$$ as the number of path connected components of $$X$$ [EDIT: Thanks to Stephen for correcting me on this]. Indeed, if $$f(1)=x$$ and $$g(1)=y$$ are two $$S^0 \to X$$ maps, any (based) homotopy between them is given by a path $$x \to y$$, and viceversa.

But! The story does not end here. In fact we can ask what happens if we get rid of the base point. Long story short, we can build something that's called a groupoid, that essentially captures both the based groups and the isomorphisms between them induce by a path from a base point to another (up to homotopy). Then $$\pi_0(X)$$ now captures information about the way to get from a point to another in the space... and that basically means to reflect the path-connected components.

Another algebraic object which gives you information about the path-connected components of the space is the zeroth homology group, $$H_0(X)$$, which is the free group on them (thus number of components = rank of $$H_0(X)$$) .

• The space $S^0$ consists of two points, and in fact there are as many basepoint-preserving maps up to homotopy from $(S^0,1)$ to $(X,x_0)$ as there are path components of $X$. – Stephen Feb 14 at 18:11
• @Stephen you're absolutely right, I made an edit. – mattecapu Feb 14 at 19:40