8
$\begingroup$

Let $X$ and $Y$ be topological spaces. Let $f: X \to Y$ be a continuous map.

Recall that $f$ is a weak homotopy equivalence iff $f$ induces group isomorphisms on the homotopy groups, i.e.: $$\forall n \geq 1: \forall > x \in X: (\hat{f}: \pi_n(X, x) \to \pi_n(Y, f(x)): [p] \mapsto [f \circ p]) \text{ is an isomorphism}$$ and $f$ induces a bijection on the set of path components, i.e: $$(\hat{f}: \pi_0(X) \to \pi_0(Y): > [p] \mapsto [f(p)]) \text{ is a bijection}$$

If we restrict the codomain to the image: $\hat{f}: X \to f(X)$ is this still a weak homotopy equivalence?

This seemed like an elementary fact to me but on closer inspection, I can't tell if it is true or not. Thank you for your help!

$\endgroup$
1
  • 1
    $\begingroup$ This is obviously not true; take $f : S^1 \to \Bbb R^2 \setminus 0$ immersed as a figure eight, one lobe wrapping about the origin. This is a homotopy equivalence, but the "corestriction" you speak of is a map $f : S^1 \to S^1 \vee S^1$, and there is no such weak homotopy equivalence. $\endgroup$ May 21, 2019 at 0:24

1 Answer 1

3
$\begingroup$

It looks to not be true. Lets call the restricted map $g:X\rightarrow f(X)$ to not overuse the $\hat f$ notation. So let $f:[0,1]\rightarrow \mathbb C$ be given by $f(x)=e^{2\pi i x}$. Since both the interval and $\mathbb C$ are contractible we have that $f$ is a weak homotopy equivalence ($\hat f:\{0\}\rightarrow\{0\}$ is an isomorphism for all $n\geq 1$ and both have one path component). Now $g:[0,1]\rightarrow f([0,1])=S^1$ given by $g(x)=f(x)$ is not a weak homotopy equivalence; take $n=1$, $x=0$, $\hat g:\pi_1([0,1],0)\rightarrow \pi_1(S^1,1)$ is a homomorphism from the trivial group to the integers, so not an isomorphism.

$\endgroup$
1
  • $\begingroup$ The $f_*$ will be mapping into $\pi_n(\mathbb C,1)$ which is why I chose the original domain to be the complex line. Having the domain of $f$ to be as "small" homotopically as possible is the trick $\endgroup$ May 21, 2019 at 0:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .