Assume $X,Y,X',Y'$ are nice spaces (not necessarily manifolds). If $\pi_{X}:X\rightarrow X'$ and $\pi_{Y}:Y\rightarrow Y'$ are local homeomorphisms, if $f':X'\rightarrow Y'$ is a continuous map, then does there exists a continuous map $f:X\rightarrow Y$ such that $f'\circ \pi_{X}=\pi_{Y}\circ f$?


  1. What about if we start of with $f$ and want an $f'$?

  2. When do we have such a property?

  3. What if $\pi_{X}$ and $\pi_{Y}$ are covering maps?


  1. If $\pi_{X}$ and $\pi_{Y}$ are bijective then yes.

  2. Somehow if the answer is yes, then $f$ must locally be $\pi_{Y}^{-1}\circ f'\circ \pi_{X}$.

  3. How to go from local property in 2. to global?

Motivation: Every continuous map $X\rightarrow Y$ induces a continuous map on its universal covers $\tilde{X}\rightarrow \tilde{Y}$

  • 1
    $\begingroup$ Do you mean $f : X \to Y$? $\endgroup$ Apr 2 at 20:08
  • $\begingroup$ @diracdeltafunk yup! $\endgroup$ Apr 2 at 20:33
  • $\begingroup$ You should correct your mistake instead of commenting "yup". $\endgroup$
    – Paul Frost
    Apr 3 at 9:20
  • $\begingroup$ @PaulFrost OK. It was a typo. The reader could have figured it out without much effort given some thought. $\endgroup$ Apr 3 at 18:48
  • $\begingroup$ @monoidaltransform Of course the reader can do it. But I do no think it is better that many readers stumble over a typo than that the OP makes a corection. $\endgroup$
    – Paul Frost
    Apr 3 at 22:45

1 Answer 1


Questions of this type are usually discussed when introducing covering spaces and various lifting properties (see for instance, Hatcher's "Algebraic Topology", chapter 1). The class of "nice" spaces consists of ones which are path-connected, locally path-connected and semilocally simply connected. (This includes all CW complexes and all topological manifolds.)

There is a difference with your question, however, since one typically works in the category of pointed spaces and asks for the existence of a lift $f$ with the prescribed values at a point. This makes answers much cleaner. For instance, if you take $X=X'=[0,1]$ and the base-points $0\in X', y'_0=f'(0)\in Y'$, then one usually asks for the existence of a lift $f$ for any choice of $y_0:=f(0)\in \pi_Y^{-1}(y'_0)$. The existence of $f$ is then called the path-lifting property. It is known that (assuming that $Y, Y'$ are nice as defined above), a local homeomorphism satisfies the path-lifting property if and only if it is a covering map. (See for instance section 5-6 "Covering spaces" in do Carmo's book "Differential geometry of curves and surfaces.")

Furthermore, for general maps of pointed "nice" spaces a lift $f$ exists (provided that $\pi_X, \pi_Y$ are covering maps) if and only if
$$ f_*(\pi_1(X,x_0))\le \pi_{Y*}(\pi_1(Y, y_0)), $$ where, $f(x_0)=y_0$.

Since you are not working in the category of pointed topological spaces, things are less predictable and I do not think there is a general theory answering your questions. One can still take $X=X'=[0,1], \pi_X=id$ and ask for a "weak" path lifting property for local homeomorphisms $Y\to Y'$ (i.e. an existence of a lift $f$ for every map $f': [0,1]\to Y'$, just as in your question). One can prove, for instance, that a finite-to-one local homeomorphism $\pi_Y: (0,1)\to S^1=Y'$ satisfies the weak path-lifting property if and only if it is a covering map. (Moreover, even locally injective maps $f': [0,1]\to S^1$ suffice to disprove the weak path-lifting property.) On the other hand, the map $$ (0,\infty)\to S^1, t\mapsto e^{it} $$ satisfies the weak path-lifting property but is not a covering map.

All in all, I suggest to stick to the usual framework of maps of pointed topological spaces.

  • $\begingroup$ This is a really great answer Moishe, it deserves more than one check given the clarity and superb exposition. $\endgroup$ Apr 5 at 22:02

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