# GP 1.2.2 $T_x(U) = T_x(X) \text{ for } x \in U.$

This is exercise 1.2.2 on Guillemin and Pollack's Differential Topology

If $U$ is an open subset of the manifold $X$, check that $$T_x(U) = T_x(X) \text{ for } x \in U.$$

I am fairly confused with this problem, because I found the implicit definition of tangent space at Guillemin and Pollack's Differential Topology comes not very handy for this problem. In the book, tangent space is defined by $d\phi_0$, which is not directly defined:

By definition, the tangent space of $X$ at $x$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.

So here's my attempt:

$T_x(X)$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.

Because $U$ is an open subset of manifold $X$, so the best linear approximation at $x$ is the same as that of $X$ around $x$. Hence, $T_x(U)$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow U$ at 0. $\phi: V \rightarrow U$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$. Therefore, $T_x(U) = T_x(X) \text{ for } x \in U.$

• "open" is important. The thing is that a local representation $\phi$ of $X$ around $x$ is, by restricting to a smaller open subset of $\mathbb{R}^k$ if necessary, also a local representation of $U$ around $x$ [and vice versa]. – Daniel Fischer Jun 25 '13 at 23:31
• In dimension 1: the tangent to the graph of $f(x)=x^2+1$ on $(-1,1)$ at $0$ has equation $y=1$. Just as if you had taken $g(x)=x^2+1$ on $\mathbb{R}$. That's the same one-dimensional affine space. Derivatives, tangent spaces, are local notion. – Julien Jun 25 '13 at 23:42
• You know how to find an equation of the tangent to the graph of $f(x)=x^2+1$ at $x=0$, right? That's $y=f(0)+f'(0)(x-0)=1$. This is exactly the same thing whether you consider $f$ defined on $(-1,1)$, or $(-37,\pi)$, or $\mathbb{R}$. The only things that matters when determining the derivative, or the tangent space, at a given point $x_0$ is the local behaviour near $x_0$. The property you ask about is the generalization of this observation. The tangent space at a point contained in an open portion of the manifold is the same as the tangent at the same point for the whole manifold, – Julien Jun 25 '13 at 23:52
• Pretty much, yes. That's what this is really about. Then to make it very rigorous, it depends on which formal definition of the tangent space (there are a few) you are using. – Julien Jun 26 '13 at 0:12
• To pass from $U$ to $X$ with this definition is trivial. You can take the exact same parametrization. In the other direction, you must say that since $U$ is an open neighborhood of $\phi(0)=x$, you may restrict $V$ if needed so that the range of $\phi$ be contained in $U$. Just by continuity of $\phi$ at $0$. – Julien Jun 26 '13 at 0:18

Let $ϕ:W∈R^k→X$ be a parametrization of $X$ around $x$ so that $ϕ(W)∈X$ is an open subset of the manifold. Since $U$ is open, so is $ϕ(W)∩U$, and also (since $ϕ$ is a homeomorphism) $V:=ϕ^{−1}(ϕ(W)∩U)$. Thus, $ϕ∣_V$ (the restriction of $ϕ$ to $V$) is a parametrization of $U$ around $x$; from this it follows directly that the tangent spaces are the same.
Let $X \subset Y$ be a submanifold and $j : X → Y$ be the inclusion map. Then $∀x \in X, dj_x : T_x(X) → T_x(Y )$ is injective—in fact it is an inclusion. If $U$ is a open subset of a manifold $X$, $T_x(U) = T_x(X)$ for $x \in U$.
Lemma: If $X$ is a manifold, $x \in X$, and $\phi : U → X$ is a local parametrization with $\phi(0) = x$, then $(d\phi_0)^{−1} = d(\phi^{−1})_x$. Now, let $x \in X$, $\phi : U → X, \psi : V → Y$ be local parameterizations $(U \subset \mathbb R^k, V \subset \mathbb R^l$ open, $\phi(0) = x = \psi(0))$. Note that $\psi^{−1} \circ j \circ \phi = \psi^{−1} \circ \phi$, and so we have $dj_x = d\psi_0 \circ d(\psi^{−1} \circ \phi)_0 \circ (d\phi_0)^{−1} = d\psi_0 \circ d(\psi^{−1})_y \circ d\phi_0 \circ (d\phi_0)^{−1} = Id$ $\blacksquare$