# Does every lift of a constant path is constant?

I'm trying to prove that every lift of a constant path is constant using the path lifting property which says that for each path $f:I\to X$ and each lift $\tilde x_0$ of the starting point $f(0)=x_0$ there is a unique path $\tilde f:I\to \tilde X$ lifting $f$ starting at $\tilde x_0$.

Maybe it's a stupid question, but I can't see the answer.

I need help.

Thanks

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Well, can you think of at least one lift of a constant path? –  Olivier Bégassat Dec 15 '12 at 9:35

Let $\pi : \tilde{X} \to X$ be the covering map. Then $\pi\circ\tilde{f} = f$.

Suppose $\tilde{f}$ is not constant. Then there are $x_1, x_2 \in I$ such that $\tilde{f}(x_1) \neq \tilde{f}(x_2)$. But as

$$\pi(\tilde{f}(x_1)) = (\pi\circ\tilde{f})(x_1) = f(x_1) = f(x_2) = (\pi\circ\tilde{f})(x_2) = \pi(\tilde{f}(x_2)),$$

$\tilde{f}(x_1)$ and $\tilde{f}(x_2)$ are in the same fibre of $\pi$. As $\tilde{f}$ is continuous and $\pi$ has discrete fibres, $\tilde{f}$ must be constant.

Added Later: The map $\pi$ is said to have discrete fibres if for every $x \in X$, the subspace topology on $\pi^{-1}(x) \subseteq \tilde{X}$ is the discrete topology (that is, every subset of $\pi^{-1}(x)$ is open). So the facts you need to know for the above argument are:

• Every covering map has discrete fibres.
• If $A$ is a connected topological space, $B$ has the discrete topology, and $g : A \to B$ is continuous, then $g$ is constant.

Note that $I$ is connected, so the second statement applies to this situation. Let me know if you need further clarification on either of these two points.

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Perhaps you can add a bit more for the OP as to why if $\tilde{f}(x_1)$ and $\tilde{f}(x_2)$ are in the same fibre of $\pi$ that $\tilde{f}$ is constant on all of $I$. –  user38268 Dec 15 '12 at 13:01
why $\pi$ has discrete fibres? by the way, what is "discrete fibres"? sorry I'm really new on this subject. Thank you for your answer. –  user42912 Dec 16 '12 at 8:48
@RafaelChavez Discrete fibre = The fibre has the discrete topology on it. This is the definition of $p$ being a covering map. –  user38268 Dec 16 '12 at 10:23

As Hatcher suggests we can deduce that the lift of a constant path is constant more directly as follows. Suppose $p : \tilde{X} \to X$ is a covering map. Then by assumption $f : I \to X$, say that it maps all of $I$ to some point $x_0 \in X$. Now because $p$ is a covering map, we have that there is an open set $U$ about $x_0$ such that $p^{-1}(U)$ is a disjoint union of open sets $\bigsqcup_{\alpha} U_\alpha$ each of which maps homeomorphically onto $U$. More concisely, we can think of $p$ as a special case of a local homeomorphism.

Now choose a lift $\tilde{f}$ of $f$ such that $\tilde{f}(0) = \tilde{x_0}$ for some $\tilde{x_0}$ in the fibre of $x_0$. Our aim now is to prove that in fact $\tilde{f} (t) = \tilde{x_0}$ for every $t \in I$. Now first of all $\tilde{x_0}$ is contained in one of those open sets (say $U_{\tilde{x_0}}$) which maps homeomorphically onto $U$. Then by continuity of the lift I can choose an open set $V$ about $0$ such that

$$\tilde{f}(V) \subseteq U_{\tilde{x_0}}.$$

Now if we apply $p$ to this and using the fact that $f$ is constant we arive at

$$p\tilde{f}(V) = x_0.$$

Now the key point now is because $\tilde{f}(V) \subseteq U_{\tilde{x_0}}$ and

$$p\bigg|_{U_{\tilde{x_0}}}: U_{\tilde{x_0}} \stackrel{\simeq}{\longrightarrow } U$$

we can apply the local inverse of $p$ to obtain that $f(V) = \tilde{x_0}$. But now this means that $f$ is locally constant on the connected set $I$ and hence is constant everywhere on $I$.

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Why you said "Then by continuity of the lift I can choose an open set V about 0 such that $\tilde{f}(V) \subseteq U_{\tilde{x_0}}$"? Thank you for your answer! –  user42912 Dec 16 '12 at 8:36
@RafaelChavez A function $f : X \to Y$ is said to be continuous if the preimage of every open set is open. Equivalently, we say that $f$ is continuous at $x \in X$ if for every $V$ open in $Y$ about $f(x)$ there is $U$ open about $X$ for which $f(U) \subseteq V$. Prove this. –  user38268 Dec 16 '12 at 10:25