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



Homotopy of paths at the boundary
If one doesn't impose any additional assumptions on $\Gamma$ or $\gamma$ then this doesn't hold.


**Counterexample.** Consider $\Gamma$ that is given by $xy(x+y)=0$, i.e. it is a union of three lines through $(0,0)$. Suppose that $\gamma\_1$ and $\gamma\_2$ are any two paths that join points $(1,0)$ and $(0,1)$ and such that their interiors don't intersect the union of lines $x=0$, $y=0$, $x+y=0$.


Now, it is not very hard to see that both of these paths can be "partially homothoped" to a path $\gamma$ that is a union of two segments lying in $xy=0$. The first segment joins $(1,0)$ with $(0,0)$ and the second $(0,0)$ with $(0,1)$.


At the same time paths $\gamma\_1$ and $\gamma\_2$ need not be homothopic in the complement in $\mathbb C^2$ to the line $x+y=0$ (since this complement has $\pi\_1=\mathbb Z$). Hence they are not partially homothopic according to your definition.


However, if **$\Gamma$ is smooth**, then the statement holds, here is an idea why.


*Idea of a proof.* The idea is that in the case $\Gamma$ is smooth we can reduce the problem to that of the problem when $\Gamma$ is the line $y=0$, when the statement is a simple exercise. The reason why this can be done is because there is a small neighbourhood of $\Gamma$, say $\Gamma\_{\varepsilon}$ that has a structure of a $2$-disk bundle over $\Gamma$. From the definition we know that both $\gamma\_1$ and $\gamma\_2$ can be homothoped inside $\Gamma\_{\varepsilon}$ so that during the homothopy the interiors of $\gamma\_1, \gamma\_2$ don't intersect $\Gamma$. In other words, we can assume from the very beginning that $\gamma\_1$ and $\gamma\_2$ are inside $\Gamma\_{\varepsilon}$.


Assume now for simplicity that $\gamma$ is itself smooth and without self-intersections (just to make it less dirty, this is not essential). Then a small neighbourhood of $\gamma$ can be mapped diffeomorphically to $\mathbb C^2$ so that $\gamma$ goes to segment $(0,0), (1,0)$, while $\Gamma$ goes to a piece of line $y=0$. Now the situation is rather clear.


