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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 $X$?

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2 Answers 2

$\pi_0$ 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.

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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
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@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
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@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.

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