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Let $p:\overline{X}\rightarrow X$ is a continuous mapping. A continuous map $s:X\rightarrow \overline{X}$ such that $p\circ s =Id_X$ is called a section of $p$. Suppose $\overline{X}$ is connected also. Suppose that $p$ is a covering projection. Show that any section of $p$ is a homeomorphism onto $\overline{X}$.

As $p\circ s=Id$ we see that $s$ is injective. Suppose we can prove that $s$ is surjective. Then we have a bijective continuous function $s$ such that $s^{-1}=p$ which means that $s$ is homeomorphism. $$p\circ s =Id\Rightarrow p\circ s\circ s^{-1}=s^{-1}\Rightarrow p=s^{-1}.$$ So, enough to prove $s$ is surjective i.e., $s(X)=\overline{X}$. As $\overline{X}$ is connected, showing $s$ is surjective is equivalent to prove $s(X)$ is both open and closed in $\overline{X}$.

Let $\bar{x}=s(x)\in \overline{X}$ for some $x\in X$.

As $(p\circ s )(x)=x$ we have $p(\bar{x})=x$. For this $x\in X$ there exists an open set $U$ in $\overline{X}$ containing $\bar{x}$ such that $U\rightarrow p(U)$ is a homeomorphism. As $U$ is open in $\overline{X}$ containing $\bar{x}$ and $s$ is continuous we have open set $V$ in $X$ with $x\in V$ and $s(V)\subset U$.

We have $$(p\circ s)(V)=V\Rightarrow (p^{-1}\circ p\circ s)(V)=p^{-1}(V)\Rightarrow (p^{-1}\circ p)(s(V))=p^{-1}(V).$$ As $s(V)\subset U$ the function $p^{-1}\circ p$ is identity on $s(V)$ so we have $s(V)=p^{-1}(V)$ So, $s(V)$ is an open set. we have $\bar{x}=s(x)\in s(V)\subset s(X)$. Thus $s(X)$ is open in $\overline{X}$.

I am not able to see why $s(X)$ is closed in $\overline{X}$.

This also says $p$ is also a homeomorphism.

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    Also, your argument is a bit shaky. You talk about the function "$p^{-1} \circ p$", but you don't know that $p$ has an inverse at that point... You can take the preimage of set under $p$, but not talk about the function $p^{-1}$. – Najib Idrissi Apr 19 '15 at 18:52
  • @NajibIdrissi On U there is an inverse.. i am taking about that inverse... –  Apr 19 '15 at 19:03
  • @NajibIdrissi : That answer has done same thing that i have done.. he said closedness can be proved similarly... So, i do not know how does that answer my question :O –  Apr 19 '15 at 19:08
  • Suppose $\hat{q} \in V \cap f(Y)$, then it follows from the same argument that $V = f(U)$ hence $\hat{p} \in V = f(U) \implies \hat{p} \in f(X)$, contradiction... – Najib Idrissi Apr 20 '15 at 07:03
  • @NajibIdrissi : Thanks –  Apr 20 '15 at 07:05

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