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. $\Box$
 A: 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.
A: 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
