# Branch point-what makes a closed loop around it special?

I am having difficulty understanding the concept of a branch point of a multifunction. It is typically explained as follows:branch point is a point such that the function is discontinuous when going around an arbitrarily small circuit around this point. What I am unable to understand is what is special about the set of points encircling this particular point? If I have a closed loop somewhere in the plane, the function is not multivalued on this loop. But if I translate this to enclose this point, then the function becomes multivalued. Why? I initially thought it might be because if this branch point is my reference point then as I run over the loop I am running over the values 0 to 2$pi$. But if I move my reference point to somewhere inside the loop this still remains true over any closed loop. Please can someone help? A similar question was asked but unfortunately did not receive ny replies. OP was advised to read the wiki page which I have.

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Perhaps the difficulty is in the definition of multifunction, rather than branch point. Have you seen a rigorous definition of this? – Zhen Lin Dec 30 '11 at 16:16
A picture is worth some $N >> 1$ number of words. Check out this visualization of the surface for $log z$. Can you see what "unfolds" at the branch cuts? upload.wikimedia.org/wikipedia/commons/4/41/… – zahbaz Aug 3 '15 at 3:42

(Disclaimer: What follows is non-standard and was dreamt up after thinking about anafunctors the whole day.)

First of all, let me define some notions. I will assume familiarity with elementary point set topology.

Definition. Let $X$ and $Y$ be topological spaces. A continuous relation between $X$ and $Y$ is a triple $(R, \sigma, \tau)$ consisting of a topological space $R$ and continuous maps $\sigma : R \to X$ and $\tau : R \to Y$.

This is a generalisation of the usual notion of a relation between two sets: indeed, recall that a relation on $X$ and $Y$ is just a subset $R \subseteq X \times Y$, and any such subset is automatically equipped with continuous projection maps to $X$ and $Y$.

Also, just as a function is a special kind of relation in set theory, a continuous function is a special kind of continuous relation: every continuous function $f : X \to Y$ induces a continuous relation $(X, \textrm{id}, f)$, and every continuous relation $(R, \sigma, \tau)$ with $\sigma : R \to X$ a homeomorphism induces a continuous function $\tau \circ \sigma^{-1} : X \to Y$.

Example. Let $X = Y = \mathbb{C}$, and let $R = \{ (x, y) \in \mathbb{C}^2 : y = x^2 \}$. Then $R$ together with the canonical projections makes a continuous relation between $\mathbb{C}$ and $\mathbb{C}$. Note that it is not a function, since the projection $(x, y) \mapsto x$ is not a homeomorphism here. (It's not even a bijection!) We will later see that it is a branched continuous multifunction, however.

Definition. An unbranched continuous multifunction $F : X \nrightarrow Y$ is a continuous relation $(R, \sigma, \tau)$ such that $\sigma$ is surjective and a local homeomorphism.

This makes $R$ into what is called an espace étalé over $X$. Here are some general facts about such spaces:

Proposition. Let $\sigma : R \to X$ be a surjective local homeomorphism.

1. For each point $x$ in $X$, the fibre $R_x = \sigma^{-1} \{ x \}$ is non-empty and has the discrete topology.

2. The map $\sigma : R \to X$ has the path lifting property: i.e. if $\gamma : [0, 1] \to X$ be a continuous path and $x = \gamma (0)$, then for any $\tilde{x}$ in $R_x$, there is a unique continuous path $\tilde{\gamma} : [0, 1] \to R$ such that $\sigma \circ \tilde{\gamma} = \gamma$ and $\tilde{\gamma}(0) = x$.

Example. Let $X = \mathbb{C} \setminus \{ 0 \}$, $Y = \mathbb{C}$, and define $$R = \{ (x, y) \in \mathbb{C}^2 : x \ne 0, y^2 = x \}$$ Let $\sigma : R \to X$, $\tau : R \to Y$ be the first and second projections, respectively. Obviously, $\sigma$ is surjective, and with a little work it can be shown that $\sigma$ is a local homeomorphism: after all, that's exactly what it means to be able to take a square root locally. Thus, $(R, \sigma, \tau)$ is an unbranched multifunction.

Now, let $\gamma : [0, 1] \to X$ be the unit circle, with $\gamma(0) = 1$. Explicitly, $$\gamma (t) = \exp (2 \pi t i)$$ One easily verifies that $(1, 1) \in R$, so by the path lifting property there is a unique path $\tilde{\gamma} : [0, 1] \to R$ lying over $\gamma$. Here we are lucky and there is an explicit formula: $$\tilde{\gamma} (t) = (\exp (2 \pi t i), \exp (\pi t i))$$

What is ‘the value’ of the multifunction along this path? It is just $\tau \circ \tilde{\gamma}$, of course. But one immediately sees that \begin{align} \tau(\tilde{\gamma}(0)) & = +1 \newline \tau(\tilde{\gamma}(1)) & = -1 \end{align} So, even though $\tilde{\gamma}$ is a lift of the closed loop $\gamma$, $\tilde{\gamma}$ itself is not a closed loop! It is precisely this which leads to the phenomenon you allude to in your question: in more advanced terms, this is simply the observation that the induced map on fundamental groups $\sigma_* : \pi_1(R, \tilde{x}) \to \pi_1(R, x)$ is not an isomorphism. When this happens, we say $\sigma$ has non-trivial monodromy.

Example. More generally, there is a notion of a covering space of $X$. If $R$ is a covering space of $X$ with covering map $\sigma : R \to X$, then $(R, \sigma, \textrm{id})$ is a unbranched continuous multifunction $X \nrightarrow R$.

Finally, let us define the notions in question themselves.

Definition. A (branched) continuous multifunction $F : X \nrightarrow Y$ is a continuous relation $(R, \sigma, \tau)$ with the following properties:

1. The map $\sigma$ is surjective.
2. The set $$U = \{ x \in X : \sigma \text{ is a local homeomorphism at each point in the fibre } R_x \}$$ is a dense open subset of $X$.
3. The restriction $(\hat{R}, \sigma |_{\hat{R}}, \tau |_{\hat{R}})$ is an unbranched continuous multifunction, where $\hat{R} = \sigma^{-1} U$.

A ramification point is a point $\tilde{x}$ of $R$ such that $\sigma$ fails to be a local homeomorphism. A branch point is a point $x$ of $R$ such that there is a ramification point $\tilde{x}$ with $\sigma(\tilde{x}) = x$. Observe that, in the above notation, $X \setminus U$ is precisely the set of branch points of $F$: so we are stipulating that the set of branch points is nowhere dense.

Proposition. If $F : X \nrightarrow Y$ is a branched continuous multifunction, and $x$ is not a branch point, then there is an open neighbourhood $U$ and a map $\phi : U \to R$ such that $\sigma \circ \phi$ is the identity on $U$, and $\tau \circ \phi : U \to Y$ is a genuine continuous function. We say $\phi$ is a local section of $F$.

This immediately follows from the fact that $\sigma$ is a local homeomorphism above $x$.

Example. Returning to the first example, where $X = Y = \mathbb{C}$ and $$R = \{ (x, y) \in \mathbb{C}^2 : y = x^2 \}$$ we see that $(R, \sigma, \tau)$ defines a branched multifunction: the only ramification point is $(0, 0)$, so the only branch point is $0$, and $U = \mathbb{C} \setminus \{ 0 \}$ is indeed an open dense subset of $\mathbb{C}$.

Now, what does this have to do with closed loops? Well, here we have to specialise a bit. One property of $\mathbb{C}$ is that it is locally simply connected: indeed, for every open neighbourhood $U$ containing a point $x$, there is a disc centred at $x$ contained in $U$. The monodromy theorem this implies that, for any surjective local homeomorphism $\sigma : R \to U$, any lift of any sufficiently small closed loop in $U$ through $\sigma$ must again be a closed loop.

Let us suppose $\sigma$ has the property that every fibre $R_x$ is finite. Thus, if $x$ is not a branch point, for every point $\tilde{x}$ in the fibre $R_x$, there is an open neighbourhood $V_{\tilde{x}}$ of $\tilde{x}$ such that $\sigma |_{V_{\tilde{x}}} : V_{\tilde{x}} \to U_{\tilde{x}}$ is a homeomorphism, where $U_{\tilde{x}}$ is an open neighbourhood of $x$. Set $$U = \bigcap_{\tilde{x} \in R_x} U_{\tilde{x}}$$ Since $R_x$ is finite, $U$ is open, and by construction $\sigma^{-1} U$ will be homeomorphic to $R_x \times U$. But $U$ must contain an open disc $D$ centred on $x$, and it is clear that $\sigma^{-1} D$ will be homeomorphic to a disjoint union of finitely many open discs. It follows that every closed loop contained in $D$ must lift to a closed loop in $\sigma^{-1} D$. Therefore, by contraposition, if $x$ is a point such that there are arbitrarily small closed loops encircling $x$ which do not lift to closed loops, $x$ must be a branch point.

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Did you mean $R=\{(x,y)\in\mathbb C^2: y^2=x \}$? – dafinguzman Oct 3 '15 at 14:57

I don't know if this will help you but here is an example.

Think about $\sqrt{z}$ going around the origin counter-clockwise, $z=re^{i\theta}$, defining $\sqrt{r}$ to be the positive square root or the non-negative real number $r$. Then it's easy to define $\sqrt{z}$ for $0\leq\theta<2\pi, r\geq0$, we have $\sqrt{re^{i\theta}}=\sqrt{r}e^{i\theta/2}$. But coming back around to the positive real axis, there is a problem. We have already defined$\sqrt{z}$ there as the positive square root of $r$, but the function, trying to continue it, wants to take the value $\sqrt{r}e^{\pi i}=-\sqrt{r}$. So you have to go to the concept of a Riemann surface or a multivalued function or whatever.

One can unambiguously define $\sqrt{z}$ on any simply connected open set not containing the origin (essentially because one can choose values for the argument of $\sqrt{z}$, a way of halving the angle). But near zero, you can't do this continuously in a single valued way as we saw above.

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