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Denote by $\Gamma$ a hypersurface in $\mathbb{C}^2$, i.e. the zero locus of a polynomial of two complex variables. Denote by $X$ the complement of $\Gamma$ in $\mathbb{C}^2$. I am trying to define a modified homotopy equivalence, namely "partial homotopy in $X$" as following: Let $\gamma_1,\gamma_2: I \rightarrow \mathbb{C}^2$ ($I$ is the unit interval) such that $\gamma_1(0)=\gamma_2(0),\gamma_1(1)=\gamma_2(1)$. We say $\gamma_1$ and $\gamma_2$ are partially homotopic in $X$ if there exists a continuous map $H: I^2 \rightarrow \mathbb{C}^2$ such that $$H(\{0\} \times I)=\gamma_i(0),H(\{1\} \times I)=\gamma_i (1) $$ $$H(t,0)=\gamma_1(t),H(t,1)=\gamma_2(t) $$ $$H(Int(I^2)) \cap \Gamma = \emptyset $$ where $Int(I^2)$ is the interior of $I^2$, i.e. $(0,1)\times (0,1)$. We write $\gamma_1 \sim_X \gamma_2$ to indicate that $\gamma_1,\gamma_2$ are partially homotopic in $X$.

Now, let $\gamma \subset \Gamma$ be a path in $\Gamma$ and let $\gamma_1,\gamma_2$ be paths such that $$\gamma_i(0)=\gamma(0),\gamma_i(1)=\gamma(1) \, , i=1,2 $$ $$\gamma_i(Int(I)) \cap \Gamma =\emptyset $$ My question is: Suppose that $\gamma_1 \sim_X \gamma, \gamma_2 \sim_X \gamma$. Does it implies that $\gamma_1 \sim_X \gamma_2$?

My attempt: It is true if I take $\gamma$ such that $\gamma((0,1]) \subset X$ by combining the homotopy as usual. By same technique, I can build a homotopy $H$ between $\gamma_1$ and $\gamma_2$ such that $$H(Int(I^2)) \cap \Gamma =\gamma$$ It is very intuitively by myself that we can lift $H$ slightly to get away from $\Gamma$ but I am lacked of topology technique to do so and I don't know to to look at. Any advice is appreciate, even modification in the hypothesis, like restrict $H$ to be a embedding, etc. Thanks

Edit: Thanks to Joshua example, this is false when we consider the union of some lines and $\gamma_1,\gamma_2$ slightly represents generators of the fundamental groups of the complements of these lines. So I think I need some extra hypothesis, like $\Gamma$ is a smooth curves in $\mathbb{C}^2$.

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First, here is a heuristic explanation of why the answer is no. The hypersurface $\Gamma \subseteq \mathbb{C}^2$ is of real codimension two, and a homotopy $H: [0,1]^2 \to \mathbb{C}^2$ will also be of real codimension two (if we choose a smooth approximation, etc. etc.). Hence, we expect them to generically intersect in real codimension four, i.e. in a $0$-dimensional submanifold of $\mathbb{C}^2$. This need not be empty. This is exactly the phenomenon that leads $\mathbb{C}^2 \setminus \Gamma$ to often have interesting fundamental groups.

Here is a counterexample: let $\Gamma = (\mathbb{C}\times 0) \cup (\mathbb{C} \times 1) \cup (0 \times \mathbb{C})$, and let $X = \mathbb{C}^2 \setminus \Gamma$. Let $\gamma_1: [0,1] \to \mathbb{C}^2$ be a loop based at $(1,0)$ such that $\gamma$ does not wind around $\mathbb{C} \times 1$ or $0 \times \mathbb{C}$, and let $\gamma_2: [0,1] \to \mathbb{C}^2$ be a loop which winds around $\mathbb{C}\times 1$ once, e.g. $$\gamma_1(t) = (1, a (1 - e^{2\pi i t})),$$ $$ \gamma_2(t) = (1, b (1-e^{2\pi i t}))$$ for $a < 1/2 < b$.

 
                          
                      Figure: The loops depicted are homotopic to
                      a loop in the hypersurface, but not each other.

Both are partially homotopic in your sense to the loop $\gamma$ which travels linearly in $\Gamma$ from $(1,0)$ to $(0,0)$ to $(0,1)$ to $(1,1)$, and then back. But $\gamma_1$ and $\gamma_2$ are not partially homotopic in $X$. Consider $Y = \mathbb{C}^2 \setminus \mathbb{C} \times 1$. If $\gamma_1$ and $\gamma_2$ were partially homotopic in $X$, then they would be homotopic relative to the basepoint $(1,0)$ in $Y$, i.e. define the same class in $\pi_1(Y,(1,0))$. But they do not, since $[\gamma_2] \in \pi_1(Y,(1,0))$ is a generator, while $0 = [\gamma_1] \in \pi_1(Y,(1,0))$.

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  • $\begingroup$ Thank you Josh. I realized that I didn't put the question exactly the way I imagined. I think my the conclusion is true when I make the assumption that $\Gamma$ is actually smooth. $\endgroup$ – Curiosity Apr 16 at 16:07

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