0

I am reading some books such as Manfredo, Lee, and Shifrin. I'm solving an example as follows. As I do not have a strong background in Topology, I am wondering few steps.

Consider a map $\gamma: \mathbb{R} \to \mathbb{R}^2$ given by $\gamma(t) = (t^3 - 4t, t^2 - 4)$. Now, given that $\alpha: (-\epsilon, \epsilon) \to \mathbb{R}$ is such that $\alpha(s) = s_0 + s$, where $p = \alpha(0)$ and $\alpha'(0) = v$, let the differential of the map $\gamma$ at $p$, denoted by $d\gamma_p$, be a map $d\gamma_p: T_p\mathbb{R} \to T_{\gamma(p)}\mathbb{R}^2$, defined by $$ d\gamma_p(v) = \frac{d}{ds}(\gamma \circ \alpha) \bigg|_{s=0}. $$ We know that the differential $d\gamma_p$ is one-to-one (I've made calculations). Clearly, $\gamma$ is surjective onto $\gamma(\mathbb{R}) \subset \mathbb{R}^2$ (as every function maps onto its image is).

The problem:

I need to decide whether $\gamma$ is a homeomorphism onto $\gamma(\mathbb{R})$.

Note: Suppose that $\gamma(\mathbb{R})$ has the induced topology from $\mathbb{R}^2$, meaning for all $W$, an open subset of $\gamma(\mathbb{R})$, we must have $W = \gamma(\mathbb{R}) \cap V$, where $V$ is an open subset of $\mathbb{R}^2$.

My attempt

I started verifying if $\gamma$ is injective, i.e., $\gamma(x) = \gamma(y)$ implies $x = y$. Thus, supposing that $$ (x^3 - 4x, x^2 - 4) = (y^3 - 4y, y^2 - 4) $$ I found that, for the second component $ x^2 - 4 = y^2 - 4 \implies x^2 = y^2 \implies x = y \text{ or } x = -y $, consequently for the first component we have $ x^3 - 4x = y^3 - 4y$ which implies that $x = y$ or $x = -y$. If $x = y$, then the function is injective however, if $x = -y$ it is not. So, at all $\gamma$ is not a homeomorphism.

My doubt:

  1. Claim: A continuous bijection from a compact set to a Hausdorff space is automatically a homeomorphism. Is this true? It would work if the domain was compact? I mean, because the differential in this case should be a tangent space over a compact space, and I never saw it. Can real subsets of a differentiable surface be compact?

  2. This second doubt is not directly related to this specific example, but it is a little bit connected with the first doubt. How can I understand that a tangent space of a differential surface doesn't contradict the definition of a tangent space over a differential manifold since a surface depends on the environment space while a differential manifold does not?

Silvinha
  • 361

1 Answers1

1

You correctly state $x^2 - 4 = y^2 - 4 \implies x^2 = y^2 \implies x = y \text{ or } x = -y$. Inserting $x = -y$ in the first component, we get $-y^3 + 4y = x^3 - 4x = y^3 - 4y$, i.e. $2y(y^2-4) = 0$. This is only possible when $y= 0$ or $y = \pm 2$. If $y = 0$, then also $x = 0$ so that $x = y$. However, we have $\gamma(2) = (0,0) = \gamma(-2)$ which shows that $\gamma$ is not injective. Therefore $\gamma$ is no embedding.

  1. You are right, a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. If the domain is not compact, the map need not be a homeomorphism. Consider for example $f : [0,2\pi) \to S^1, f(t) = e^{it}$.
    I do not which definition of "surface" you are using, but if you understand it as $2$-manifold, then it can of course be compact (for example, the $2$-sphere). If you require it to be a graph of as smooth map defined on an open subset of the plane, then it cannot be compact.

  2. There is an abstract definition of the tangent space for general manifolds and a "geometric" definition for surfaces in $\mathbb R^3$. Both are equivalent. See for example tangent vector of manifold.

Kritiker der Elche
  • 5,474