Let us have a topological space $X$. What is the relation between path-homotopy classes and path-components? For example, can we somehow define a map from path-components to path-homotopy classes?
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What exactly do you mean by "path-homotopy classes"? – xyzzyz May 16 '16 at 18:21
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A path-homotopy class is written as $[f]$ such that it consists of paths $g$ in $X$ that are path-homotopic to $f$ – Ninja May 16 '16 at 18:32
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Really, path components are concerned with the homotopy classes of base-pointed maps $\{0,1\}\to X$ (or non-base-pointed maps $\{0\}\to X$). If $0$ is the basepoint of $S^0=\{0,1\}$ then there is a 1-1 correspondence between $\pi_0(X)=[(S^0,0),(X,x_0)]_*$ and the set of path components of $X$, given by bijectively associating to a path component $C$ of $X$ the class of maps $[f]$ in $\pi_0(X)$ such that $f(1)\in C$.
The set of all classes of path-homotopic paths is only loosely related to the set of path components of a space as the usual definition has at least one class of paths for every pair of points in the same path component $C$.
One could quotient out by the relation which is the transitive closure of the relation $[f]\sim[g]$ if $f(1)=g(0)$. Then $$\{\mbox{classes of path-homotopic paths}\}/{\sim^t} \longleftrightarrow \{\mbox{path components}\}$$
is a 1-1 correspondence, but this doesn't seem very natural as you would have already had to have known the $\pi_0$ equivalence class of $f|_{\{0,1\}}$ for all the $f$ with $f(0)=x_0$ to calculate the above equivalence relation, so then why not just stick with $\pi_0$?
Dan Rust
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