# Modified homotopy and relation with intersection theory.

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 $$\mathbb{C}^2$$.

## 1 Answer

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 $$0 = [\gamma_1] \in \pi_1(Y,(1,0))$$.

• 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. – Curiosity Apr 16 at 16:07