# Corestricting a weak homotopy equivalance

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!

• 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. May 21, 2019 at 0:24

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.
• 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 May 21, 2019 at 0:29