# I Cauchy's Theorem

1. May 22, 2017

### Silviu

Hello! I am reading a book on complex analysis and I came across this: If $G \in \mathbb{C}$ is a region, a function f is holomorphic in G and $\gamma$ is a piecewise smooth path with $\gamma \sim_G 0$ then $\int_\gamma f = 0$. I want to make sure I understand. First of all, $\gamma \sim_G 0$ means that if G doesn't have "holes", any closed loop is homotopic to a point? And this also means that if G doesn't have holes, the integral of any holomorphic function over a closed loop is 0? Which means that in G any holomorphic function has an antiderivative? Thank you!

2. May 22, 2017

### FactChecker

What it specifically says is that the particular path $\gamma$ can be shrunk to a point within G. It doesn't say that all paths can be shrunk to a point within G.
(But in the special case of G being simply connected, any closed path can be shrunk to a point within G.)
Right.
Right.

3. May 23, 2017

### Silviu

Thank you! One more question, as far as I remember a circle and a point are not topologically equivalent as they have a different number of holes. So topologically a circle can't be shrunk to a point. But homotopically it can. What is the difference between the two? In the case of a homotopy you ignore the holes? (I understand mathematically the definition of homotopy and homeomorphism I am just not sure I can visualize the difference from a geometrical point of view).

4. May 23, 2017

### Staff: Mentor

A point and a circle are not homeomorph since they are of different dimensions. You will lose information. Or formally, you cannot establish a bijection. A circle and a point, if regarded as the graphs of two functions are homotopic, since there is a continuous function (the shrinking) that transforms one into the other. However, if you cut out a point in the inner area of the circle, they won't be homotopic anymore, since the "shrinking" would have to "jump" across that whole, i.e. it cannot be done continuously anymore.

5. May 23, 2017

### FactChecker

Topology is not my strength, but here is a thought about terminology.
We should distinguish between "shrink to a point" and "shrink to become a point". The first would mean that, given any open set containing the point, the path becomes completely contained in that open set. The second would mean that there is a homotopy between the line and a point, which is not possible(?).
Perhaps there is already better terminology than "shrinks to a point".

6. May 23, 2017

### Staff: Mentor

It is possible: $H : \mathbb{S}^1 \times [0,1] \longrightarrow \{0\}\, , \,H(\varphi ,t)=(1-t)\cdot \begin{bmatrix}\cos \varphi \\ \sin \varphi \end{bmatrix}$ is continuous, $\{H(\varphi,0)\,\vert \,\varphi \in [0,2\pi)\} = \mathbb{S}^1\; \textrm{ and } \; \{H(\varphi ,1)\,\vert \,\varphi \in [0,2\pi)\} = \{(0,0)\}$ is a homotopy that shrinks the circle.

7. May 28, 2017

### WWGD

By cardinality reasons alone, a circle and a point are not topologically equivalent/ homeomorphic.

8. May 29, 2017

### lavinia

A space that can be shrunk to a point is called "contractible". This means that it can be shrunk to a point inside itself. That is: there is a homotopy i.e. a continuous map $H:M×[0,1]→M$ such that at time $0$ $H$ is the identity map and at time $1$ it projects the space $M$ onto a point $p$ in $M$.

An example of a contractible space is Euclidean space. The continuous map $H(x,t) = (1-t)x$ is the identity at time $0$ and maps everything to the origin at time $1$. Another example is a set of line segments that share a common end point.

If the topological space $M$ is a subset of another space $N$ then one says that $M$ is null homotopic in $N$ if it can be shrunk to a point in $N$. In this case the homotopy maps $M×[0,1]$ into $N$. That is: $H: M×[0,1]→N$ is the inclusion of $M$ in $N$ at time $0$ and maps $M$ to a point in $N$ at time $1$.

A circle is not contractible. There is no homotopy $H:S^1×[0,1]→S^1$ that is the identity at time $0$ and projects the circle onto one of its points at time $1$.

But a circle inside Euclidean space is homotopic to a point since the same homotopy that shrinks all of Euclidean space to a point shrinks any subset to a point. That is: every subset of Euclidean space is null homotopic.

In complex analysis one considers regions of the plane that are not contractible. Typical examples are contractible open sets - such as a disk - minus a finite number of points or minus a finite number of disks, for instance an annulus. Within such regions, there are closed curves that are not null homotopic - for instance in a disk minus a point, a closed curve that winds around the missing point a finite number of times. Also there are closed curves that are null homotopic - for instance any closed curve that does not wind around the missing point. None of these closed curves can be shrunk to a point in themselves but some of them can be shrunk to a point within the region.

Observations:

- Two spaces are said to be homotopically equivalent if there are continuous maps $H:M →N$ and $G:N→M$ such that $HG$ and $GH$ are both homotopic to the identity map. Homotopically equivalent is not the same as homeomorphic. A contractible space is homotopically equivalent to a point but not in general, homeomorphic to a point. An annulus is homtopically equivalent to a circle.

- In some spaces, every closed curve is null homotopic. Such a space is said to be "simply connected". For instance, the sphere of any dimension greater than one, is simply connected. A simply connected space need not be contractible though. No sphere is contractible.

- In Algebraic Topology one measures the failure of a space to be simply connected by its fundamental group. By definition, a space is simply connected if its fundamental group is trivial. The fundamental group of a point is trivial. But the fundamental group of a circle is the integers.

- If two spaces are homotopically equivalent, then they have isomorphic fundamental groups. Since Eulcidean space is homotopically equivalent to a point, its fundamental group is trivial. A circle is not homotopically equivalent to a point since its fundamental group is not trivial.

Last edited: May 29, 2017