Recall a weak homotopy equivalence is a continuous map $f:X_1 \rightarrow X_2$ that induces a bijection $f_*:[Y,X_1] \rightarrow [Y,X_2]$ for any CW-complex $Y$, where $[Y,X]$ denotes the homotopy class of continuous maps $Y \rightarrow X$. When $Y$ is restricted to be $\Bbb{S}^n$, this results in a series of isomorphisms between $\pi_n(X_1)$ and $\pi_n(X_2)$, and by Whitehead theorem it automatically becomes a homotopy equivalence when $X_i$'s are CW-complexes.
Why don't we define it in a similar way as homotopy equivalence, namely $(f_*)^{-1}:[Y,X_2] \rightarrow [Y,X_1]$ is induced by another continuous map $g:X_2 \rightarrow X_1$? Does the latter one necessarily implies homotopy equivalence? What will the difference of the two definitions change if we restrict $Y$ to be $\Bbb{S}^n$?


2 Answers 2


The reason why the definition of weak homotopy equivalence is good is because of Whitehead's theorem. It is the weakest definition (meaning the easiest to check) that is equivalent to homotopy equivalence for CW complexes, so you would not want to change it.

As to your new question, the Hawaiian earring is a counter example. There is a map $X \rightarrow H$ that is a weak homotopy equivalence from a cw complex to $H$, but there can be no inverse that you ask for because the fundamental group of $H$ has nontrivial topology while the fundamental group of $X$ is discrete (as it is for any CW complex).

  • $\begingroup$ I think my new question is: If a inverse that induces $(f_*)^{-1}$ actually exists, does it imply homotopy equivalence? $\endgroup$
    – Zerox
    Commented Nov 10, 2021 at 14:32
  • 1
    $\begingroup$ No, look up the double comb space. It is weakly equivalent to a point and so satisfies the inverse existing, but it is not Homotopy equivalent to a point. $\endgroup$ Commented Nov 10, 2021 at 15:06
  • $\begingroup$ You must distinguish between pointed and unpointed homotopy classes. As it stands, with the notation given in the question, the definition is not that of a weak equivalence. See for instance example 1.2. $\endgroup$
    – Tyrone
    Commented Nov 10, 2021 at 19:37

EDIT: Now that the question has been modified, this answer is no longer directly relevant.

The standard definition of weak homotopy equivalence is a map $f: X_1 \to X_2$ such that the induced maps on homotopy groups, for all choices of base point in $X_1$, are isomorphisms; see https://ncatlab.org/nlab/show/weak+homotopy+equivalence or https://en.wikipedia.org/wiki/Weak_equivalence_(homotopy_theory) or p. 352 in Hatcher. Your definition is stronger, and I think that it is equivalent to a homotopy equivalence: if $f_*: [Y, X_1] \to [Y, X_2]$ is a bijection for all $Y$, let $Y=X_2$; then there is a map $g \in [X_2,X_1]$ which maps to the identity on $X_2$, and that ought to be a homotopy inverse to $f$.

  • $\begingroup$ Well, I lost some restriction on $Y$. I'll add it. $\endgroup$
    – Zerox
    Commented Nov 9, 2021 at 14:05

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