A topological space $X$ is $n$-connected if the homotopy groups $\pi_r(X)$ for $0 \leq r \leq n$ are trivial groups. This means (let's say geometrically), $X$ is $0$-connected if it is non-empty and path-connected. It is $1$-connected if it is simply connected.
Now, Let X be a topological space and A⊂X a subspace. For x∈A, the homotopy group $\pi_n(X,A,x)$ of the pair $(X,A)$ is by definition the set of homotopy classes of n-cells relative to X modulo A (the usual definition). Now, a pair is $n$-connected if $\pi_r(X,A)$ is trivial for $0 \leq r \leq n$.
Is there any interpretation (geometrical) of $0$-connectedness and $1$-connectedness of a pair $(X,A)$? In general what does it mean (geometrically) by an $n$-connected pair?
Best Answer
An element $f:(D^n,S^n,s_0)\to(X,A,x_0)$ in $\pi_n(X,A,x_0)$ is zero if it is homotopic through such maps to the constant map to $x_0$. Precomposing the null-homotopy with a homotopy $D^n\to D^n\times I$ rel $\partial D^n$ starting at $D^n\times\{0\}$ and ending at $D^n\times\{1\}\cup \partial D^n\times I$, we have a homotopy rel $\partial D^n$ to a map $(D^n,s_0)\to (A,x_0)$. On the other hand, given a map $(D^n,s_0)\to (A,x_0)$, there is a null-homotopy in $A$ since $D^n$ deformation retracts to $s_0$. That means a map $f:(D^n,S^n,s_0)\to(X,A,x_0)$ represents $0$ iff $f$ is homotopic rel boundary to a map $D^n\to A$.
If $\pi_n(X,A,x_0)=0$ for all $x_0\in A$, then every map $(D^n,∂D^n)\to (X,A)$ is homotopic rel $∂D^n$ to a map to $A$, and vice versa. Note that if $A$ is path-connected, then there is a isomorphism $\beta_\gamma:\pi_n(X,A,x_0)\cong\pi_n(X,A,x_1)$ for any path $\gamma:x_0\to x_1\in A$, that means the relative homotopy group does not depend on the basepoint and is then written as $\pi_n(X,A)$.
The set $\pi_1(X,A,x_0)$ is not a group but rather a pointed set, with the basepoint $0$ referring to the homotopy class of the constant path $x_0$. In this dimension, $\beta_\gamma:\pi_1(X,A,x_0)\to\pi_1(X,A,x_1)$ is still a bijection even if $\gamma$ is not a path in $A$, but if want $\beta_\gamma$ to be a pointed map, we should still require $\gamma$ to be in $A$.
We could try to extend the definition to $n=0$. Since $D^0=\{0\}$ and $∂D^0=\emptyset$, an element in $\pi_0(X,A,x_0)$ would be simply a homotopy class of a point in $X$, i.e. a path component, and $0$ would be the path component containing $x_0$. However, using the characterization of the zero element as the class of a map $D^0\to X$ which is homotopic rel $∂D^0$ to a map to $A$, we could interpret any point in $X$ as $0$ as long as it can be connected via a path to a point in $A$, in other words, every path component intersecting $A$ is regarded trivial. This suggest that we define $\pi_0(X,A,x_0)$ is the quotient of $\pi_0(X,x_0)$ by the set $i_*(\pi_0(A,x_0))$, where $i_*$ is induced by the inclusion $i:A\to X$.
With this definition, a $0$-connected pair $(X,A)$ is one with $i_*:\pi_0(A,x_0)\to\pi_0(X,x_0)$ being surjective, meaning that $A$ intersects every path component of $X$.