Example: closed path not homotopic to path in subset I am looking for an example for the following setting:
Given an open subset $U$ of $M$, both path-connected, such that there is a closed path in $M$ that is not homotopic to a closed path in $U$, while $M/U$ is simply connected.  
Explanations:


*

*$M/U$ shall denote the quotient identifying all points of $U$ to a single point, formally $m_1\sim m_2$ iff either $m_1=m_2$ or both $m_1,m_2\in U$.

*We may fix a base point $u\in U$. Then a closed path in $M$ is a continuous map $\gamma:[0,1]\to M$ with $\gamma(0)=\gamma(1)=u$.


Just in case it is possible I would appreciate if $M$ is a manifold.
Thanks in advance.
 A: After writing all the stuff below this I realized there is a simpler example which is even a manifold. But since it took some effort I'm not deleting it :)
Let $M = S^1$ and $U = S^1 \setminus \{ z \}$ where $z$ is not the base point. The identity $\gamma : S^1 \to S^1$ is not homotopic to a path in $U$, for otherwise it would be nullhomotopic. And the quotient is $M/U$ is the Sierpiński space which is contractible.
Note that you're basically asking an example of a pair of path-connected spaces $(M,U)$ where $\pi_1(M/U) = *$ but $\pi_1(M,U) \neq *$. If one of $M$ or $U$ were simply connected and the inclusion were a cofibration then this can't happen due to the Blakers–Massey theorem. In the example above it fails because $U \subset M$ is not a cofibration. There is probably an example where the inclusion is a cofibration but both $M$ and $U$ aren't simply connected.

Here is a counterexample (which is not a manifold however). Let $M$ be the pseudocircle: it has four elements, $M = \{a,b,c,d\}$, and the open sets are $\{\{a,b,c,d\},\{a,b,c\},\{a,b,d\},\{a,b\},\{a\},\{b\},\emptyset \}$. Take as open subset $U = \{a,b,c\}$ and base point $a$.
The quotient space is $M/U = \{x, d\}$ ($x = [a] = [b] = [c]$) with topology $\{\{x,d\}, \{x\}, \emptyset\}$. This is the Sierpiński space, which is contractible (thus simply connected). 
Define a loop
$$\begin{align}
\gamma : [0,1] & \to M, \\
t & \mapsto
\begin{cases}
a & \text{if } 0 \le t < 1/3, \\
b & \text{if } t = 1/3, \\
c & \text{if } 1/3 < t < 2/3, \\
d & \text{if } t = 2/3, \\
a & \text{if } 2/3 < t \le 1.
\end{cases}
\end{align}$$
As explained in the Wikipedia article, this loop defines a weak homotopy equivalence $S^1 \to M$, so it is not nullhomotopic. Therefore it cannot be homotopic to a path in $U$, because $U$ is contractible.
