# Proof verification: weak homotopy equivalences and homotopy equivalences

Let $$X := \{0\} \cup \{\frac{1}{n} \ | \ n \in \mathbb{N}\}$$ be the topological space with the subspace topology. Regard $$\mathbb{N}$$ as a topological space with the discrete topology. Show that there is a weak homotopy equivalence $$\mathbb{N} \to X$$, but that it is not a homotopy equivalence. Moreover, show that there is no weak homotopy equivalence $$X \to \mathbb{N}$$.

I have a faulty proof for the last part. That probably also means my proof for the first part is wrong, so I will post it here as well.

As each point in $$\mathbb{N}$$ is a path component, its fundamental groups $$\pi_n(\mathbb{N}, n_0)$$ are trivial for all $$n \geq 1$$, $$n_0 \in \mathbb{N}$$. Each element $$\frac{1}{n}$$ is a connected component, so also a path component. As connected components partition $$X$$, $$0$$ must be a connected component as well, hence a path component. By the same logic, $$\pi_n(X, x_0)$$ is trivial for all $$n \geq 1$$ and $$x_0 \in X$$. That means that any map $$\mathbb{N} \to X$$ is a weak homotopy equivalence. Any function is continuous, so take for example $$f(0) = 0, f(n) = \frac{1}{n}$$. Let $$g$$ be a homotopy inverse of $$f$$, then $$f \circ g \simeq \textrm{id}_\mathbb{X}$$. This homotopy defines a paths between $$f(g(\frac{1}{n}))$$ and $$\frac{1}{n}$$. That must mean that $$g(\frac{1}{n}) = n$$ and $$g(0) = 0$$. But $$\frac{1}{n} \to 0$$, while $$g(\frac{1}{n})$$ does not converge to $$0$$, which means that $$g$$ is not continuous, contradiction.

Now for the second part, suppose $$f: X \to \mathbb{N}$$ is a continuous map. As all homotopy groups are trivial, that means it is a weak homotopy equivalence. THere is certainly a continuous map, for example constant maps.

• Note that not all of the homotopy groups of either $X$ or $\mathbb{N}$ are trivial - there is $\pi_0$. Jan 10, 2020 at 19:38
• @JasonDeVito thanks, so the first one should still work because $f$ is also a bijection between path components. For the second part, is this valid? Let $f: X \to \mathbb{N}$ be weak homotopy equivalence. In particular it is a continuous bijection. Then $f^{-1}(f(0))$ is open, so it contains $0$ and infinitely many $\frac{1}{n}$, but then $f$ is not injective, contradiction. Jan 10, 2020 at 20:06
• @William good point. How would we then prove that $\{0\}$ is a path component? Intuitively it makes sense to me, but I find it hard to give a rigorous argument. Jan 10, 2020 at 20:07
• Suppose $\gamma\colon I\to X$ is continuous and $\gamma(t) = 1/n$ for some $n$ and some $t$. Since $\{1/n\}$ is both open and closed, so is $\gamma^{-1}(1/n)$; since this pre-image is non-empty (it at least contains $t$), it must be all of $I$ because $I$ is connected. Therefore any path containing $0$ must be constant, because if it also passed through some $1/n$ then it would be constantly $1/n$ which is a contradiction. Jan 10, 2020 at 20:15
• @Pel: My initial reaction was that "In particular it is a continuous bijection" is wrong because that's often not the case for homotopy equivalences. But a bit of thinking convinced me it is right in the case. Perhaps its better to spell that out in more detail? (What I was thinking for a proof: Show that any continuous $f:X\rightarrow \mathbb{N}$ has finite image, which, in particular, implies that the induced map on $\pi_0$ fails to be surjective.) Jan 10, 2020 at 20:53

Your proof of the first part is essentially correct, but you forgot $$\pi_0$$. This requires to show that all path components of $$X$$ are singletons which is easy to prove. And now you see that you cannot take any $$\phi : \mathbb N \to X$$. In order that $$\phi$$ be a weak homotopy equivalence, it must establish a bijection between the path components of both spaces. Since these are singletons, $$\phi$$ must be a bijection. However, any bijection will do.
For the second part let us see what it means that a function $$f : X \to \mathbb N$$ is continuous.
By continuity there exists $$\varepsilon > 0$$ such that $$\lvert f(x) - f(0) \rvert < 1$$ for $$\lvert x - 0 \rvert < \varepsilon$$. This implies that $$f(x) = f(0)$$ for $$\lvert x - 0 \rvert < \varepsilon$$. Thus if $$n > 1/\varepsilon$$, then $$f(1/n) = f(0)$$. This shows that $$f$$ takes only finitely many values (which are contained in the set $$\{f(0)\} \cup \{f(1/n)\mid n \le 1/\varepsilon \}$$).
Hence $$\pi_0(f) : \pi_0(X,x_0) \to \pi_0(\mathbb N,f(x_0))$$ is never a bijection (for any $$x_0$$).
By the way, you can use this also to show that no bijection $$\phi : \mathbb N \to X$$ can be a homotopy equivalence. If it were one, then it would have a homotopy inverse $$g$$. This map would also be a homotopy equivalence, in particular it would give us a bijection $$\pi_0(g) : \pi_0(X,0) \to \pi_0(\mathbb N,g(0))$$. But this is impossible as we have seen.
• $\mathbb{N}$ has the discrete topology, so isn't a map $f: X \to \mathbb{N}$ continuous if and only if $f^{-1}(n)$ is open for all $n \in \mathbb{N}$? Your proof still works because $f^{-1}(f(0))$ is open, which means it must contain some $[0, \epsilon[$ which contains almost all of $X$. Anyway, thank you, I understand now. Jan 12, 2020 at 15:56
• You are right, using the fact that $f^{.-1}(f(0))$ is open is simpler. But the conclusion is the same: All but finitely points of $X$ mist be mapped to a single point of $\mathbb N$. Jan 12, 2020 at 22:33