# Zeroth Homotopy intuition

I have two questions regarding the notion of zeroth homotopy of a space, any help would be greatly appreciated.

The first is about the zeroth homotopy of a connected space. In my teachers notes on the subject it says that "$\mathbb{R}^n$ is path connected [...] thus $\pi_0(\mathbb{R}^n)$ has one point". However whilst reading around to try and understand $\pi_0$ a little better I have come across theorems that say that "$\pi_0(X)=0$ iff X is a connected space". To me these two statements contradict each other but I am unsure if it is a difference in convention or I am misunderstanding something? Are path connected spaces not connected?

Secondly, I have an exercise that asks me to "Calculate $\pi_0$ of the following spaces: $(\mathbb{R}\setminus\{0\})$, $S^2$, $\mathbb{Q}$,...". For a question like this, is it usually expected to just write down the cardinality of the set $\pi_0$ for the space? Or is there a more detailed way to describe $\pi_0$?

Many thanks

• connected spaces are not always path-connected. – Henno Brandsma Oct 31 '17 at 12:44

It is just a difference in convention. Here $0$ stands for the trivial one-point zeroth homotopy. In additive group notation, $\{0\}$ is the one-element group.

It also looks like you're correct that the exercise wants you to find the cardinality of the zeroth homotopy classes for the spaces. This should work out easily if you know how many path-connected components each space has.

• That's great, thanks for your help – Munkres Oct 31 '17 at 12:30
• You need the path-components instead. – Henno Brandsma Oct 31 '17 at 17:44
• Yes, whoops! Mix those up all the time. They happen the same for the three sets given, right? – Andrew Tindall Oct 31 '17 at 18:42
• For these sets yes, they are the same. – Henno Brandsma Oct 31 '17 at 22:49

The $0$-th homotopy "group" $\pi_0(X,b_0)$ ($b_0$ is the base point) is just the set of equivalence classes (under base-point fixing homotopies) of base-point respecting continuous functions from $S^0 = (\{-1,1\},-1)$ to $(X, b_0)$.
(In this case all functions are continuous on the discrete set $S^0$ regardless of $X$ and so such maps $f$ are just determined by the value $f(1) \in X$, as $f(-1) = b_0$).
The maps $f$ with $f(1) = x$ and $g$ with $g(1)= y$ are homotopic iff there is a continuous path $p: [0,1] \to X$ from $x$ to $y$. (Easy exercise in definitions)
So $f$ is equivalent to $g$ iff $f(1)$ and $g(1)$ lie in the same path-component. So $\pi(X,b_0)$ is just the set of path-components of $X$, essentially.
So a connected but not path-connected space like the topologist's sine curve has a two point $\pi_0$. There is no natural group structure on this, as with loops (where we have an extra composition of loops operation). The cardinality is the only real information. A singleton says that the space is path-connected, as is the case for your most of your example spaces. In $\mathbb{Q}$ the path-components, like the components, are its singletons, so $\pi_0(\mathbb{Q})$ is countable.