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.

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    $\begingroup$ Note that not all of the homotopy groups of either $X$ or $\mathbb{N}$ are trivial - there is $\pi_0$. $\endgroup$ Jan 10, 2020 at 19:38
  • $\begingroup$ @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. $\endgroup$
    – user388557
    Jan 10, 2020 at 20:06
  • $\begingroup$ @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. $\endgroup$
    – user388557
    Jan 10, 2020 at 20:07
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    $\begingroup$ 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. $\endgroup$
    – William
    Jan 10, 2020 at 20:15
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    $\begingroup$ @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.) $\endgroup$ Jan 10, 2020 at 20:53

1 Answer 1


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.

  • $\begingroup$ $\mathbb{N}$ 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. $\endgroup$
    – user388557
    Jan 12, 2020 at 15:56
  • $\begingroup$ 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 $\mathbb N$. $\endgroup$
    – Paul Frost
    Jan 12, 2020 at 22:33

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