Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $X$ be a topological space. we define an equivalence class on $X$ by $x\sim y$ if there exists a path $\gamma:I\to X$ that joins $x$ to $y$. now the zeroth homotopy set is the quotient $\pi_0(X)=X/_\sim$. My question is why we call it a set isn't it a topological space with the quotient topology induced from $X$?

share|cite|improve this question

$\pi_0$ is the set of path-components of $X$, and of course you can give it the quotient topology. But for a very large class of spaces path components are open and so the quotient topolgy $\pi_0$ will be discrete.

share|cite|improve this answer
you mean the following: if $q:X\to \pi_0(X)$ is the quotient map and $[x]$ denotes the equivalence class of some $x\in X$ then for many spaces we have that $q^{-1}([x])$ is open in $X$ and so by definition of quotient topology we must have $[x]$ open in $\pi_0(X)$ which means that this quotient topology is discrete. is this what you mean? and why is that path components are open for many spaces? – palio Feb 23 '12 at 12:39
@palio: Any manifold, simplicial complex or cell complex is locally path connected, so path components are open. These form a very large subset of spaces studied by people. – Grumpy Parsnip Feb 23 '12 at 12:46
that's exactly what i don't see.. why being locally path connected implies path components are open?? – palio Feb 23 '12 at 12:57
@palio: By definition, every point in a path component has an open neighborhood which is path connected, so that means this neighborhood is contained in the path component. So the path component is open. – Grumpy Parsnip Feb 23 '12 at 13:10

Actually you are counting the path connected components of your topological space, what is important is the cardinality of this set rather than its topology.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.