To show that the complement of the embedded interval in R^2 is path connected Let $c:I\to\mathbb{R}^2$ be continuous injective curve.
How can I show that $\mathbb{R}^2-{\rm im}(c)$ is path connected?
I first guessed and tried to show that pair $(\mathbb{R}^2,{\rm im}(c))$ is homeomorphic(or homotopic)
to ($\mathbb{R}^2$, linear segment) But I'm not sure that this is correct and even if this
is true, it seems to tough for me.
Next I tried to use homology theory, some mayer vietoris sequence but I'm not
making any progress now. I would appreciate your help.
 A: You are correct that for any embedding of a closed interval $c\subset \mathbb R^2$, the pair $(\mathbb R^2,c )$ is homeomorphic to $(\mathbb R^2, I)$ where $I$ is standardly embedded. See this math.overflow question that addresses this. As you suspect, it is a difficult statement to prove. 
My idea to apply Mayer Vietoris was to let $U$ be the complement of the curve and let $V$ be an $\epsilon$ neighborhood of $c$, but I don't see how to easily argue that $U\cap V$ is connected, which is what you'd need for this argument to work.
The easiest way I see to do the problem requires the machinery of Alexander Duality: if $X$ is a compact, locally contractible subspace of $S^n$, then $\widetilde{H}_q(S^n\setminus X)\cong \widetilde{H}^{n-q-1}(X),$ where $\widetilde{H}$ stands for reduced homology and cohomology. So in our case $\widetilde{H}_0(S^2\setminus c)\cong H^1(c)=0$. So $S^2\setminus c$ is path connected. Removing a point from $S^2$ yields $\mathbb R^2$. Since it is easy to arrange for a path to miss a point, $\mathbb R^2\setminus c$ is connected as well.
