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Let $p:\overline{X}\rightarrow X$ be a simply connected covering of a path connected space $X$ and $A\subset X$ be a path connected set.

Show that the inclusion induced homomorphism $i_{\sharp} : \pi_1(A)\rightarrow \pi_1(X)$ is surjective iff $p^{-1}(A)$ is path connected.

Assume that $i_{\sharp} $ is surjective. Let $\bar{x},\bar{y}\in p^{-1}(A)$ then we have $p(\bar{x}),p(\bar{y})\in A$.

As $A$ is path connected we have a path $\omega : I\rightarrow A$ such that $\omega(0)=p(\bar{x})$ and $\omega(1)=p(\bar{y})$.

As $p$ is a covering projection we can lift this to a path in $\overline{X}$

There exists $\eta : I\rightarrow \overline{X}$ such that $p\circ \eta =\omega$. So, we have $p(\eta(I))=\omega(I)\subset A\Rightarrow \eta(I)\subset p^{-1}(A)$.

So, we see that the lift is actually in $p^{-1}(A)$ we also know that $\eta(0)=\bar{x}$ but it is not clear why should $\eta(1)=\bar{y}$.

I have used just that $A$ is path connected and nothing else...

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Using the surjectivity of $\pi_1(A,x_0)\to\pi_1(X,x_0)$, one can show that every path in $X$ between two points in $A$ is homotopic to some path in $A$. This this end, let $x$ and $y$ be points in $A$. There is a path $\alpha$ from $x$ to $y$ in $A$. Let $\beta$ be any path from $x$ to $y$ in $X$. Then $\beta \simeq \lambda\cdot\alpha$ for some loop $\lambda$ at $x$, and this $\lambda$ is homotopic to some $\kappa$ in $A$. Therefore, $\beta$ is homotopic to $\kappa\cdot\alpha$ in $A$.

Now let $\tilde x$ and $\tilde y$ be two points in $p^{-1}(A)$ with $x=p(\tilde x)$ and $y=p(\tilde y)$. Since $\tilde X$ is path connected, there is a path $\sigma$ from $\tilde x$ to $\tilde y$. Then $p\sigma$ is a path from $x$ to $y$, and thus homotopic to some path $\omega$ in $A$. Since $p$ is a covering map, their respective lifts $\sigma$ and $\eta$ at $\tilde x$ must be homotopic as well, and that means that $\eta$, a path in $p^{-1}(A)$, ends at the same point as $\sigma$, namely $\tilde y$.

The simply-connectedness of $\tilde X$ is only relevant for the proof of the other direction.

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  • $\begingroup$ Hi... Thanks for your answer... i have done the converse also.. see if there are some gaps.. suppose that $p^{-1}(A)$ is path connected. Let $a\in A\subset X$ and $\omega$ be a loop in $X$ based at $a$. This can be lifted to a path $\bar{\omega}$ in $\overline{X}$. As $p^{-1}(A)$ is path connected we have a path $\tau : I\rightarrow p^{-1}(A)$ with $\tau(0)=\bar{\omega}(0)$ and $\tau(1)=\bar{\omega}(1)$. Compose with $p$. We have $p\circ \tau : I \rightarrow A$. $\endgroup$ – user87543 Apr 19 '15 at 10:05
  • $\begingroup$ So, we hae two paths $\bar{\omega} : I\rightarrow \overline{X}$ and $\tau : I\rightarrow \overline{X} $ with $\tau(0)=\bar{\omega}(0)$ and $\tau(1)=\bar{\omega}(1)$. As $\overline{X}$ is simply connected these two paths has to be homotopic. So, we have $\bar{\omega}\approx \tau$ which implies $p\circ \bar{\omega}\approx p\circ \tau$ i.e., $\omega \approx p\circ \tau$. As $\tau (I)\subset p^{-1}(A)$ we have $ (p\circ \tau)\subset A$. So, we have a loop $p\circ \tau$ in $A$ based at $a$ such that $\omega\approx p\circ \tau$. Thus $i_{\sharp}$ is surjective. $\endgroup$ – user87543 Apr 19 '15 at 10:06
  • $\begingroup$ @PraphullaKoushik: Yes, everything is done right. $\endgroup$ – Stefan Hamcke Apr 19 '15 at 11:49

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