# How to show that a chain is homologous to zero

Hey my first question here, so I got an exercise but need some help. The questions is:

I have a domain $$U\subset\mathbb{C}$$, and curves $$\gamma_j : [0, 1] \to U$$, $$j \in \{0, 1\}$$ which are continuously differentiable.

They have the same start and end point, so $$\gamma_0(0) = \gamma_1(0) = z_0$$ and $$\gamma_0(1) = \gamma_1(1) = z_1$$. There is also a Homotopy between these points, $$H : [0, 1] \times [0, 1] \to U$$ with $$H(0, \cdot) = \gamma_0$$, $$H(1, \cdot) = \gamma_1$$, $$H(\cdot, 0) = z_0$$ and $$H(\cdot, 1) = z_1$$.

Show $$\Gamma = \gamma_0 - \gamma_1$$ is homologous to zero.

Here is the definition of homologous to zero:

If $$\Gamma$$ is in $$U$$ and if $$Int(\Gamma) \subset U$$, then $$\Gamma$$ is called homologous to zero. With $$Int(\Gamma) = \{z\in\mathbb{C}\setminus[\Gamma]\mid ind_\Gamma(z) \neq 0\}$$. $$Int(\Gamma)$$ is called Interior and $$ind_\Gamma(z)$$ is called winding number.

I probably have to show that $$ind_{\gamma_0}(z) = ind_{\gamma_1}(z) \ \forall z \notin U$$. Which probably implies $$Int(\gamma_0 - \gamma_1) = Int(\Gamma) \subset U$$ and therefore $$\Gamma$$ is homologous to zero.

Any ideas how to do this? Thanks in advance!

EDIT:

I think I got something but maybe there are some arguments missing:

Proof:

Let $$z\in\mathbb{C} \setminus U$$. Since $$H$$ is continuously differentiable, the mapping $$j \mapsto ind_{\gamma_j}(z) = \frac{1}{2\pi i}\int_{\gamma_j} \frac{1}{\xi - z}d\xi = \frac{1}{2\pi i}\int_0^1 \frac{1}{H(j, t) - z}\frac{\partial^2 H(j,t)}{\partial j \partial t} dt$$ is continuously from $$[0, 1]$$ to $$\mathbb{R}$$. Also, the values from this map are in $$\mathbb{Z}$$. Therefore, the mapping is constant.

This implies $$ind_{\gamma_0}(z) = ind_{\gamma_1}(z) \ \forall z \notin U$$.

Which implies $$Int(\gamma_0 - \gamma_1) = Int(\Gamma) \subset U$$ and therefore $$\Gamma$$ is homologous to zero. $$\Box$$

• I think the phrase you're looking for is "homologous to zero." – D. Brogan Jun 22 at 12:20
• Your question does not make much sense. How is the subtraction $\gamma_0 - \gamma_1$ defined? How is $Int(\Gamma)$ defined? And "Is $\Gamma$ in $U$ and is $Int(\Gamma) \subset U$, so $\Gamma$ is called null-homolog" is not even a grammatical sentence, let alone a reasonable definition of what is called "homologous to zero" in English. – Lee Mosher Jun 22 at 12:48
• @D.Brogan thank! I'll have a look maybe I'll find something regarding this :) – Xvid Jun 22 at 12:56
• @LeeMosher sorry, english isn't my first language so I'm sorry for grammatical errors. Regarding your questions, we never explicitly defined $\gamma_0 - \gamma_1$. $Int(\Gamma)$ is an open subset, I have defined it in my post. Well, besides the grammatical error this is unfortunately the defintion my professor gave us. *I edited the post now, hopefully it's grammatically now a bit more correct :) – Xvid Jun 22 at 12:56
• @Xvid Your proof doesn't work. See my above comment. I recommend to have a look at math.stackexchange.com/q/112679. – Paul Frost Jun 23 at 12:28

The discussion in the comments shows that some concepts need to be clarified. Here is a pragmatic approach.

You start with curves $$\gamma_j : [0, 1] \to U$$, $$j \in \{0, 1\}$$, which are continuously differentiable. Let us assume that the homotopy $$H : I \times I \to U$$ from $$\gamma_0$$ to $$\gamma_1$$ is a continuous map which is partially differentiable with respect to the second coordinate (i.e. $$\dfrac{\partial H}{\partial s}(t,s)$$ exists for all $$(t,s) \in I \times I$$) and such that $$\dfrac{\partial H}{\partial s} : I \times I \to \mathbb C$$ is continuous. This implies that all intermediate curves $$H_t = H(t,-)$$, $$t \in I$$, are continuously differentiable. This requirement is a little bit weaker than assuming that $$H$$ is continuously differentiable (this concept would need further clarification).

Let us moreover agree that the winding number $$ind_\gamma(z)$$ is defined for closed piecewise continuously differentiable curves $$\gamma$$ and all $$z \notin \gamma(I)$$. The winding number is always an integer given by $$ind_\gamma(z) = \frac{1}{2\pi i} \int_\gamma \frac{1}{\zeta-z}d\zeta = \frac{1}{2\pi i} \int_0^1 \frac{\gamma'(s)}{\gamma(s)-z}ds .$$

Finally we regard $$\Gamma = \gamma_0 - \gamma_1$$ as a closed curve based at $$z_0$$. It is composed by $$\gamma_0$$ (path from $$z_0$$ to $$z_1$$) followed by the inverse of $$\gamma_1$$ (path from $$z_1$$ to $$z_0$$). Explicitly, $$\Gamma(s) = \begin{cases} \gamma_0(2s) & s \le 1/2 \\ \gamma_1(2-2s) & s \ge 1/2 \end{cases} \quad$$ .

Let us define a homotopy $$G : I \times I \to U, G(t,s) = \begin{cases} H(t,s) & s \le 1/2 \\ \gamma_1(2-2s) & s \ge 1/2 \end{cases} \quad$$ . This is a homotopy from $$G_0 = \Gamma$$ to $$G_1 = \gamma_1 - \gamma_1$$ such that all $$G_t$$ are piecewise continuously differentiable curves.

By definition $$G(I\times I) \subset U$$. It therefore suffices to show that for all $$z \in \mathbb C \setminus G(I\times I)$$ we have $$ind_\Gamma(z) = 0$$ because this implies that $$Int(\Gamma) \subset G(I\times I) \subset U$$.

For $$z \in \mathbb C \setminus G(I\times I)$$ we obviously have $$ind_{G_1}(z) = \frac{1}{2\pi i} \int_{\gamma_1} \frac{1}{\zeta-z}d\zeta - \frac{1}{2\pi i} \int_{\gamma_1} \frac{1}{\zeta-z}d\zeta = 0$$. We want to prove that $$ind_\Gamma(z) = ind_{G_0}(z) = 0$$. To do this, it has to be shown that the function $$\phi : I \to \mathbb Z, \phi(t) = \frac{1}{2\pi i} \int_0^1 \frac{G_t'(s)}{G_t(s)-z}ds$$ is continuous and therefore constant. I shall not give a proof but leave it as an excercise to you.

Thanks to everyone, I'll provide the solution from my teacher which seems, at least for me, a bit more easier to understand.

We have to show that $$Int(\Gamma) = Int(\gamma_0 - \gamma_1) \subset U$$, or in other words $$ind_{\Gamma}(z) = ind_{\gamma_0 - \gamma_1}(z) = 0 \ \forall z \notin U$$.

Let us define the mapping

$$\eta_s(z) : [0, 1] \to \mathbb{R}, s \mapsto ind_{\gamma_0 - \gamma_s}(z) \ \forall z \notin U$$

$$\rightarrow ind_{\gamma_0 - \gamma_s}(z) = \frac{1}{2 \pi i} \int_{\gamma_0 - \gamma_s} \frac{d\zeta}{\zeta - z}= \frac{1}{2 \pi i} \int_{\gamma_0} \frac{d\zeta}{\zeta - z} - \frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z}$$

The first term is obviously constant, as he doesn't vary. The second term can we rewrite as:

$$- \frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z} = - \frac{1}{2 \pi i} \int_0^1 \frac{\gamma_s^\prime(t)}{\gamma_s(t) - z} dt.$$

We know, from some previous Proposition from our lecture, that this is continuous and $$\in \mathbb{Z}$$, therefore this one is constant too.

$$\Rightarrow \eta_s(z) = const.$$

We only have to calculate now what the value of $$\eta_s(z)$$ for every $$s\in [0,1]$$ is (as it's constant).

Let $$s = 0 \rightarrow \eta_0(z) = ind_{\gamma_0 - \gamma_0}(z) = 0$$.

Therefore, $$\eta_1(z) = ind_{\gamma_0 - \gamma_1}(z) = ind_{\Gamma}(z) = 0 \ \forall z \notin U$$. $$\Rightarrow Int(\Gamma) \subset U$$ and therefore $$\Gamma$$ is homologous to zero.

With this, one can also easily come to the homotpy version of Cauchy's integral theorem.

It's probably not a perfect proof but this kinda works out for me, sorry for grammar errors

• The proof is not correct. You consider the function $I(s) = -\frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z} = - \frac{1}{2 \pi i} \int_0^1 \frac{\gamma_s^\prime(t)}{\gamma_s(t) - z} dt$ and claim that it is continuous (which is certainly true) and an integer which is not true. An integral of the form $\frac{1}{2 \pi i} \int_{\gamma} \frac{d\zeta}{\zeta - z}$ only takes integer values for closed curves. – Paul Frost Jun 24 at 16:54
• However, you can easily repair it. Although $I(s) = -\frac{1}{2 \pi i} \int_{\gamma_s} \frac{d\zeta}{\zeta - z}$ doesn't take integer values, it is continuous and therefore also $\eta_s$ is continuous. But $\eta_s$ takes integer values and thus is constant. – Paul Frost Jun 24 at 17:14
• @PaulFrost thanks! Maybe I missed this argument but yeah make sense. – Xvid Jun 25 at 9:06