simply connected covering of a path connected space (II)

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 injective iff each path component of $p^{-1}(A)$ is simply connected.

First of all i do not think i understand what does it mean to say $i_{\sharp} : \pi_1(A)\rightarrow \pi_1(X)$ is injective..

Is it like if i have a loop in $A$ which is not nullhomotopic in $A$ (existence of $H:I\times I\rightarrow A$) then it is not nullhomotopic (existence of $H:I\times I\rightarrow X$) in $X$.

Let me know if this is what it actually mean...

Assume $i_{\sharp}$ is injective and let $\omega$ be a loop in $p^{-1}(A)$... See this as a loop in $\overline{X}$... as $\overline{X}$ is simply connected $\omega$ is nullhomotopic in $\overline{X}$ i.e., we have $H:I\times I\rightarrow \overline{X}$ such that

$H(t,o)=\omega(t)$ and $H(t,1)=w(0)$ for all $t\in I$..

Compose this with $p$ to get $I\times I\xrightarrow{p\circ H}X$ with $(p\circ H)(t,0)=(p\circ \omega)(t)$ and $(p\circ H)(t,1)=(p\circ \omega)(0)$..

So, this $p\circ \omega$ is null homotopic in $X$ so it has to be null homotopic in $A$ as well.. As $p\circ \omega$ is nullhomotopic in $A$ i belive this would imply $\omega$ is nullhomotopic in $p^{-1}(A)$..

I could not think of any ideas about converse part and how to prove if $p^{-1}(A)$ is not actually path connected...

You are right about the first part. If $p\omega$ is nullhomotopic in $A$, then the homotopy to the constant loop lifts to a homotopy between $\omega$ and the constant loop. This is because of the lifting properties of covering maps. Note that a covering map $p:\tilde X\to X$ restricts to a covering map $p^{-1}(A)\to A$ for every subset $A$ of $X$.

Regarding the second part, let every path component of $p^{-1}(A)$ be simply-connected. If $\omega$ is a loop in $A$, nullhomotopic as a loop in $X$, then what can you say about its lift $\tilde\omega$. Recall that every homotopy of paths lifts to a homotopy between their unique lifts at a given point $\tilde x$ lifting the starting point of the paths.

• Let $\omega$ be a loop in $A$ null as a loop in $X$... We have $H:I\times I \rightarrow X$ such that $H(t,0)=\omega(t)$ and $H(t,1)=\omega(0)$. We then have a lift $\tilde H : I\times I \rightarrow \tilde X$ such that $p\circ \tilde H= H$... The lift $\tilde\omega$ may not be a loop in $\tilde X$.. we get a path in $p^{-1}(A)$.. As $a=\tilde\omega(0)$ may not be same as $b=\tilde\omega(1)$ we can get a path $\tau$ from $a$ to $b$ then $\tau*\tau^{-1}$ is a loop in $p^{-1}(A)$ and as $p^{-1}(A)$ is simply connected we have $\tau*\tau^{-1}\approx c$..I do not know how to proceed next – user87543 Apr 23 '15 at 20:31
• @PraphullaKoushik: The important observation is that the lift $\tilde\omega$ is a loop. The loop $\omega$ is homotopic as a path to the constant loop $c$. This homotopy lifts to a homotopy $\tilde H$ between $\tilde\omega$ and $\tilde c$. But $\tilde c$ is constant, thus $c(1)=c(0)$ and, since $\tilde H$ is a homotopy of paths, $\tilde\omega(s)=c(s)$ for $s=0,1$. – Stefan Hamcke Apr 23 '15 at 20:55
• Oh.. Ok ok.. So we actually have a loop but then...what next.. As that lift is a loop it has to be nullhomotopic.. Then compose with p again and say that this is nullhomotopic in A... – user87543 Apr 23 '15 at 21:14
• @PraphullaKoushik: Yes, the lift is a loop in $p^{-1}(A)$, thus nullhomotopic since the path components of that set are simply-connected. From this we can conclude that $\omega$ is nullhomotopic in $A$. – Stefan Hamcke Apr 23 '15 at 22:35
• Got it... Thank you :) – user87543 Apr 24 '15 at 2:59

Since the question asks only for hints I refer to the following paper (available here)

R. Brown, P/R. Heath, and K.H. Kamps, Groupoids and the Mayer-Vietoris sequence'', J. Pure Appl. Alg. 30 (1983) 109-129.

which deals with the more general question of components and vertex groups of the pullback by a morphism $f: A \to C$ of a groupoid fibration or covering of a groupoid $C$.

The relevance to coverings of spaces is explained in Topology and Groupoids Chapter 10.

• I thank you for your reply but I am afraid I am not fully ready to read a paper at this time as I am just learning the subject... I will definitely come to this once I am ready.. Thank you – user87543 Apr 23 '15 at 21:17

The Statement is not true. Für example $X=\overline{X}={\mathbb R}^2$, which is simply connected, p the Identity, and $A=S^1$. The Fundamental Group of A is nontrivial, so $i_\sharp$ is not injective, but $p^{-1}A=A$ is connected.

By the way, the Statement as above has no meaning anyway. The Path component is path connected anyway, you must mean something else.

• I am sorry, I mean to say it is simply connected... – user87543 Apr 23 '15 at 19:23