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I want to prove:

$X$ is simply connected so every continuous function $f:X\to\mathbb{C}^*$ is homotopic to a constant function.

Now what I've noticed is that for example $S^1$ is a deformation retract of $\mathbb{C}^*$ but I don't know if that helps. I'm a bit stuck with that, any ideas please?

Thomas Andrews
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Zaragosa
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    Try lifting $f$ to the universal cover of $\mathbb{C}^{\ast}$. – Thorgott Oct 15 '22 at 00:07
  • @Thorgott I can take the universal cover $exp:\mathbb C\to \mathbb C^*$. But after that I don't know what to use. Would you give me some reference to read a theorem to help me here? I know how to build paths but $f$ is not a path, or I am seeing something wrong. – Zaragosa Oct 15 '22 at 00:45
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    You want to use a theorem that if $X$ is simply connected and $Y$ has a universal cover $\widetilde{Y}$ then every continuous map $f : X \to Y$ lifts to a continuous map $\widetilde{f} : X \to \widetilde{Y}$. Do you know it? – Qiaochu Yuan Oct 15 '22 at 00:53
  • @QiaochuYuan I did not know that theorem, where can I find it so I can read about it? – Zaragosa Oct 15 '22 at 00:55
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    @Zaragosa, probably in every single topology textbook that deals with covering spaces. – Mariano Suárez-Álvarez Oct 15 '22 at 01:32
  • @MarianoSuárez-Álvarez I have the Munkres and Hatcher at hand and I don't see it, I think I'm confusing something. If anyone can give me a reference I would greatly appreciate it. – Zaragosa Oct 15 '22 at 01:41
  • You are very right! I haven't gotten to that chapter yet... now I'll start reading. Let's assume for a moment that I already know this theorem, so I would have a continuous function $\tilde{f}:X\to\mathbb C$ such that $exp\circ \tilde{f}=f$. I define the continuous mapping $H(x,t)=exp(t\cdot \tilde{f}(x))$ such that $H(x,0)=exp(0)=(1,0)=e_{ (1,0)}$ and $H(x,1)=exp(\tilde{f}(x))=f(x)$. What do you think? – Zaragosa Oct 15 '22 at 02:16

1 Answers1

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It is not true for all simply connected $X$. A counterexample will be given later.

You are right, $S^1$ is a strong deformation of $\mathbb C^*$, thus your statement is equivalent to

If $X$ is simply connected, then every continuous function $f:X \to S^1$ is homotopic to a constant function.

But let us ignore this. You know that $\exp : \mathbb C \to \mathbb C^*$ is a universal cover. Therefore we can apply the well-known lifting theorem for covering maps - but only under the additional assumption that $X$ is locally path connected. In that case we get

  • If $X$ is simply connected and locally path connected, each $f : X \to \mathbb C^*$ has a lift $\tilde f : X \to \mathbb C$ (i.e. $\exp \circ \tilde f = f$).

Since $\mathbb C$ is contractible, $\tilde f$ is homotopic to a constant function, and therefore the same is true for $f = \exp \circ \tilde f$.

Let us now give the promised counterexample.

The Warsaw circle is a simply connected compact subset of $\mathbb C^*$. See for example To show that Warsaw circle is simply connected. Let us choose a copy $W \subset \mathbb C$ of the Warsaw circle containg $0$ in its interior. We may assume that the line segment $L = \{-1 + iy \mid 1 \le y \le 3 \}$ is the subset of $W$ where the oscillating $\sin$-curve part accumulates.

We claim that the inclusion $i : W \hookrightarrow \mathbb C^*$ is not homotopic to a constant function.

Assume to the contrary that there is a homotopy $H : W \times I \to \mathbb C^*$ such that $H_0 = i$ and $H_1$ is a constant function with value $c \in \mathbb C^*$.

By the Tietze extension theorem the map $$G : \bar W = W \times I \cup \mathbb C \times \{1\} \to \mathbb C, G(z,t) = \begin{cases} H(z,t) & z \in W \\ c & t = 1 \end{cases}$$ has a continuous extension $K : \mathbb C \times I \to \mathbb C$ (extend the two coordinate functions $\bar W \to \mathbb R$ of $G$ to get $K$). The set $U = K^{-1}(\mathbb C^*)$ is an open neigborhood of $\bar W$ in $\mathbb C \times I$. By the tube lemma we find an open neighborhood $V$ of $W$ in $\mathbb C$ such that $V \times I \subset U$. Then $j = K_0 \mid_V : V \to \mathbb C^*$ is a continuous extension of $i$ which is homotopic in $\mathbb C^*$ to the constant map with value $c$.

$V$ contains an open disk $B_r(z_0)$ with center $z_0 = -1 + i$ and radius $r$ which is mapped by $j$ into the set open disk $B_1(z_0) \subset \mathbb C^*$. Pick a point of the form $z_1 = a + i \in W$ such that $-1 < a < -1 +r$. Then the line segment $L'$ connecting $z_0$ and $z_1$ is contained in $V$. Remove $L$ and the part of the oscillating $\sin$-curve between $L$ and the vertical line through $a$ and insert $L'$ into this "gap". This gives a compact subspace $W' \subset U$ which is a topogical copy of $S^1$. Clearly the map $i' = j \mid_{W'} : W' \to \mathbb C^*$ is is homotopic in $\mathbb C^*$ to the constant map with value $c$.

Take a homeomorphism $h : S^1 \to W'$ by travelling counterclockwise around $W'$. Then $\phi = r \circ i' \circ h : S^1 \to S^1$ is homotopic to a constant map, where $r : \mathbb C^* \to S^1$ denotes the standard retraction. But by construction $\phi$ is homotopic to the identity which gives a contradiction.

Paul Frost
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