A pair $(X,A)$ is $n$-connected iff the inclusion $A\rightarrow X$ is an $n$-equivalence.

Question

This book says followings:

Definition. A continuous map $e:A\rightarrow Z$ is an n-equivalence if, for all $y\in Y$, the induced map $e_{*}:\pi_q(Y,y)\rightarrow \pi_q(Z,e(y))$ is an injection for $q< n$ and a surjection for $q\leq n$.

Definition. A pair $(X,A)$ is said to be $n$-connected if $i_*:\pi_0(A)\rightarrow \pi_0(X)$ is surjective and $\pi_q(X,A,a)=0$ for $1\leq q\leq n$ and all $a\in A$, where $i:A\rightarrow X$ is the inclusion. It is equivalent that the inclusion $i:A\rightarrow X$ be an $n$-equivalence.

My problem is the last statement. I guess it is not trivial. How do I prove it? Is this immediately induced by the definition?

Answer 1

Use the long exact sequence of homotopy groups for $(X,A,a)$.

Let us first consider $n = 0$. Note that the pointed set $\pi_0(Z,z)$ denotes the set $\pi_0(Z)$ of all path components of $Z$ with basepoint the path component $P_Z(z)$ of $z \in Z$.

$i$ being a $0$-equivalence means that $i_* : \pi_0(A,a) \to \pi_0(X,a)$ is surjective for all $a \in A$. This is the same as $i_* : \pi_0(A) \to \pi_0(X)$ being surjective. This is the definition of $(X,A)$ $0$-connected.

Let us now consider $n > 0$.

1. Let $(X,A)$ be $n$-connected. Then you get for $1 \le q < n$ an exact sequence $$0 \to \pi_q(A,a) \to \pi_q(X,a) \to 0$$ which means that $i_*$ is an isomorphism in these dimensions. For $q = n$ you get $$\pi_n(A,a) \to \pi_n(X,a) \to 0$$ which means that $i_*$ is a surjection in dimension $n$. For $q = 0$ you get $$0 \to \pi_0(A,a) \to \pi_0(X,a)$$ which shows that only the path component of $a$ in $A$ is contained in the path component of $a$ in $X$. Since this is true for all $a \in A$, we see that $i_* : \pi_0(A) \to \pi_0(X)$ is injective. By definition of "$n$-connected" it is also surjective.
   Hence $i$ is an $n$-equivalence.

2. Let $i$ be an $n$-equivalence. This means that it is also a $0$-equivalence, and above we have seen that $i_* : \pi_0(A) \to \pi_0(X)$ is surjective in this case. For $1 \le q \le n$ consider the exact sequence $$\pi_q(A,a) \stackrel{\alpha_q = i_*}{\to} \pi_q(X,a) \stackrel{\beta_q}{\to} \pi_q(X,A,a) \stackrel{\partial}{\to} \pi_{q-1}(A,a) \stackrel{\alpha_{q-1} = i_*}{\to} \pi_{q-1}(X,a)$$ $\alpha_q$ is surjective and $\alpha_{q-1}$ injective. Thus $\ker \beta_q = \operatorname{im} \alpha_q = \pi_q(X,a)$ which means that $\operatorname{im} \beta_q = 0$. Moreover, $\operatorname{im} \partial = \ker \alpha_{q-1} = 0$, thus $\ker \partial = \pi_q(X,A,a)$. We get $\pi_q(X,A,a) = \ker \partial = \operatorname{im} \beta_q = 0$. Note that for $q=1$ the last three terms are no groups, but only pointed sets. This does not affect the above argument.