Weak homotopy equivalence of topological spaces

This article defines a property of continuous maps between topological spaces

Definition

Definition for path-connected spaces in terms of homotopy groups

Let $A$ and $B$ be path-connected spaces. A weak homotopy equivalence from $A$ to $B$ is a continuous map $f:A\to B$ such that the functorially induced maps $\pi _{n}(f):\pi _{n}(A)\to \pi _{n}(B)$ are group isomorphisms for all $n\geq 1$ .

Note that since the maps are homomorphisms anyway, it is enough to require them to be bijective.

Basepoint choice disclaimer for homotopy group isomorphism: To concretely define the map $\pi _{n}(f)$ , we need to choose basepoints for $A$ and $B$ . Change of basepoint, however, only results in pre-composition and post-composition by isomorphisms and does not affect whether or not the map is an isomorphism.

Definition for spaces that are not path-connected

Let $A$ and $B$ be topological spaces. A weak homotopy equivalence from $A$ to $B$ is a continuous map $f:A\to B$ such that:

The functorially induced map $\pi _{0}(f):\pi _{0}(A)\to \pi _{0}(B)$ is a bijection between the set of path components $\pi _{0}(A)$ and the set of path components $\pi _{0}(B)$ .
For every path component of $A$ , the restriction of $f$ to a continuous map from that to its image path component of $B$ is a weak homotopy equivalence of path-connected spaces.

Facts

The existence of a weak homotopy equivalence from $A$ to $B$ does not imply the existence of a weak homotopy equivalence from $B$ to $A$ . Thus, to get an equivalence relation on topological spaces, we need to take a symmetric transitive closure. We say that spaces are weak homotopy-equivalent topological spaces if they are in the same equivalence class under the equivalence relation thus obtained.
The mere fact that $\pi _{n}(A)\cong \pi _{n}(B)$ as abstract groups is not enough to guarantee that $A$ and $B$ are weak homotopy-equivalent, even when $A$ and $B$ are manifolds or CW-spaces (see isomorphic homotopy groups not implies weak homotopy-equivalent). Rather, it is specifically important that the map must induce those isomorphisms.
The exception to the above is in the case that both $\pi _{n}(A)$ and $\pi _{n}(B)$ are the trivial group/one-point set for all $n$ . In this case, any map must induce isomorphisms since that's the only possible map between trivial groups/one-point sets. In this case, the spaces $A$ and $B$ are both weakly contractible spaces.
Similarly, the mere fact that $\pi _{1}(A)\cong \pi _{1}(B)$ as abstract groups and $H_{n}(A)\cong H_{n}(B)$ as abstract groups does not imply that $A$ and $B$ are weak homotopy-equivalent. See isomorphic homology groups and isomorphic fundamental groups not implies weak homotopy-equivalent. Rather, it is specifically important that the map must induce those isomorphisms.
The exception to the above is, once again, where the fundamental group and all the homology groups $H_{n},n\geq 1$ , are trivial.