Meaning of n-connected pairs 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? 
 A: Here are two facts:


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*Any $n$-connected pair $(X,A)$ can be approximated by CW pairs $(X',A')\to (X,A)$ inducing isomorphisms on all the relative and absolute homotopy groups. 

*A CW pair $(X',A')$ is $n$-connected iff up to homotopy equivalence of the pair, the cells of $X'-A'$ have dimensions $>n$. 
Intuitively, $n$-connected pairs $(X,A)$, up to weak homotopy equivlance (or homotopy equivalence if restricted to CW pairs), are such that $A$ "contains all dimension $\le n$ parts" of $X$.
A: Hatcher's Algebraic Topology offers a very good (and rigorously proven) set of algebraic topological theorems for this exact problem. Read Section 4.1. The real content of what you are looking for starts on page 354. His text can be downloaded for free here. He looks at the case where $A$ is a non-empty CW complex. The upshot is that arbitrarily constructed n-connected CW modules end up being unique up to homotopy equivalence. Propositions 4.17 and 4.18 highlight this point.
A: 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 $X$.
