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Any continuous map $f: X_1 \to X_2$ "lifts" to a map $\tilde f: \tilde X_1 \to \tilde X_2$ (provided that $X_1$ and $X_2$ have universal covers).

The space $\tilde X_1$ is certainly path-connected and locally path-connected and we have $$(f\circ p_1)_*(\pi_1(\tilde X_1))=\{0\}=(p_2)_*(\pi_1(\tilde X_2)).$$ Then the lifting criterion ensures the existence of a lift $\ \tilde f: \tilde X_1 \to \tilde X_2$. Is this a very basic fact I somehow forgot, or have I made some glaring error?

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    $\begingroup$ Certainly one can lift in this manner. I'm not sure I understand what the question is. $\endgroup$ – user98602 Aug 28 '14 at 15:18
  • $\begingroup$ @MikeMiller: Thanks for the sanity check. My question was simply whether or not the statement (and my justification) was correct. I suppose I just never saw any applications of this fact when I first learned about covering spaces. $\endgroup$ – Kyle Aug 28 '14 at 15:21
  • $\begingroup$ No problem. I don't think I've seen an application of this at all; if you have one in mind, I'd love to hear it. $\endgroup$ – user98602 Aug 28 '14 at 15:23
  • $\begingroup$ @MikeMiller: It was used by studiosus here in the context of lifting a homeomorphism $f: X\to X$ to a homeomorphism $\tilde f: \tilde X \to \tilde X$. $\endgroup$ – Kyle Aug 28 '14 at 15:27

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