If $q: E\rightarrow X$ is a covering map that has a section (i.e. $f: X\rightarrow E, q\circ f=Id_X$) does that imply that $E$ is a $1$-fold cover?
If a covering map has a section, is it a $1$-fold cover?
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general-topology
algebraic-topology
covering-spaces
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1Does it have *only one* section? – 2012-12-12
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2Well, what if $E = X \amalg X$? – 2012-12-12
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0@ZhenLin: I forgot to add that $E$ has to be connected...because that obviously would not hold in case $E$ is not connected, as you pointed out. – 2012-12-12
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0@Andy I'm not sure how that makes a difference? – 2012-12-12
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0Well, think about $\mathbb{R}$ covering $S^1$, or $\mathbb{C}\setminus \{ 0 \}$ covering itself with the map $z \mapsto z^n$. – 2012-12-12
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0Addressing the other question by OP in the answers, I would like to suggest the following: $f\colon x \mapsto (x,0)$ and $q\colon(x,y)\mapsto x$ – 2012-12-12
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0For $p$ to have a section, does $p$ have to be a homeomorphism? – 2012-12-12