If I take the torus $T$ of revolution in $\mathbb{R}^3$ is very intuitive for me that there is an open subset of $T$ that is diffeomorphic to a cylinder, for example $T$ minus a circle on the $xz$ plane. But how can I prove this in a rigourous way? This problem appeared to me when I learn how to calculate the cohomology groups of the deRham complex of $T$, where is assumed that we can take such a type of open subsets, and we use those open sets in the Mayer-Vietoris Sequence. Can anyone give me a help? Thanks.
Why the torus has open subsets diffeomorphic to a cylinder
algebraic-topologygeneral-topologygroup-cohomologyhomology-cohomologymayer-vietoris-sequence
Related Solutions
Jim's comment is right, the glueing information is hidden in the map $\mathbb R^2 \to \mathbb R^2$ corresponding to the map $\phi : H^2_c(U\cap V) \to H^2_c(U) \oplus H^2_c(V)$.
Recall that if $U$ and $V$ are opens of $\mathbb R^2$, if we have a map $\iota : \mathbb U \to \mathbb V$, it induces a map $\iota^* : \Omega^2_c(\mathbb U) \to \Omega^2_c(\mathbb V)$, such that $\iota^* (f dxdy) = g dxdy$ where $g$ satisfies $f = (g \circ \iota) * J$ where $J$ is the jacobian of $\iota$, and $g=0$ outside the image of $\iota$. Consequently, depending on the (non-changing) sign of $J$, we have $\int_U \omega = \pm \int_V (\iota^*(\omega))$ forall $\omega \in \Omega^2_c(U)$ (the change of variable formula is exactly what we have, but with an absolute value on the jacobian). Therefore you have to keep track wether all your inclusion maps preserve orientation or not.
Write $U \cap V = W = W_1 \cup W_2$, so that $H^2_c(U\cap V) = H^2_c(W_1) \oplus H^2_c(W_2)$. We know that $U,V,W_1,W_2$ are diffeomorphic to $\mathbb R^2$, so the isomorphisms are induced by $\alpha : \omega \in \Omega^2_c(U) \mapsto \int_U \omega$
$\phi$ is given by $\phi(\omega_1 \oplus \omega_2) = (\iota_{W_1 \to U}^*(\omega_1) - \iota_{W_2 \to U}^*(\omega_2), \iota_{W_1 \to V}^*(\omega_1) - \iota_{W_2 \to V}^*(\omega_2))$.
So $\alpha \circ \phi (\omega_1 \oplus \omega_2) = (\int_U \iota_{W_1 \to U}^*(\omega_1) - \int_U \iota_{W_2 \to U}^*(\omega_2), \int_V \iota_{W_1 \to V}^*(\omega_1) - \int_V \iota_{W_2 \to V}^*(\omega_2))$.
In the Möbius case, we usually pick maps such that $\iota_{W_1 \to U},\iota_{W_2 \to U},\iota_{W_1 \to V}$ are orientation preserving, and $\iota_{W_2 \to V}$ is orientation-reversing, so we obtain :
$\alpha \circ \phi (\omega_1 \oplus \omega_2) =
(\int_{W_1} \omega_1 + \int_{W_2} \omega_2, \int_{W_1} \omega_1 - \int_{W_2} \omega_2)$
thus the corresponding map $\tilde{\phi} : \mathbb R^2 \to \mathbb R^2$ is $(x,y) \mapsto (x+y,x-y)$, which is an isomorphism. Therefore, its kernel and cokernel, $H^1_c(M)$ and $H^2_c(M)$, are both zero.
We cut $T^{3}$ into two parts, each part is homotopic to a torus. One visualize this by considering $T^{3}=T^{2}\times \mathbb{S}^{1}$, and the two parts are $\mathbb{T}^{2}\times I$ respectively, with the $I$ coming out of considering $\mathbb{S}^{1}$ as gluing two intervals together. The intersection of the two parts is again homotopic to the torus. Nowe we have:
$$\rightarrow H^{2}(X)\rightarrow H^{2}(\mathbb{T}^{2}\times I)\oplus H^{2}(\mathbb{T}^{2}\times I)\rightarrow H^{2}(\mathbb{T}^{2}\times I)\rightarrow H^{3}(X)\rightarrow0$$
We know $H^{2}(\mathbb{T}^{2})=\mathbb{R}^{1}$ via induction. So we have
$$H^{0}(X)\rightarrow \mathbb{R}^{2}\rightarrow \mathbb{R}\rightarrow H^{1}(X)\rightarrow \mathbb{R}^{4}\rightarrow \mathbb{R}^{2}\rightarrow H^{2}(X)\rightarrow \mathbb{R}^{2}\rightarrow \mathbb{R}\rightarrow \mathbb{R}\rightarrow0$$
This gives $H^{2}(X)=\mathbb{R}^{3}$ because last map is an isomorphism and the map $H^{1}(\mathbb{T}^{2}\times I)\rightarrow H^{2}(X)$ has a one dimensional image. Consider a closed one-form $w$ on $\mathbb{T}^{2}$, if we use partition of unity to split it into two parts, then no matter which choice we use we would end up with the same class in $H^{2}(X)$ if one thinks geometrically. This together with the first part gives us $H^{1}(X)=\mathbb{R}^{3}$, $H^{2}(X)=\mathbb{R}^{3}$.
Best Answer
It depends on your definition of the torus. For instance, if the torus is $S^1 \times S^1$ then you might remove $\{x_0\} \times S^1$.
We know $(S^1 \times S^1) \setminus (\{x_0\} \times S^1)$ is the same thing as $(S^1 \setminus \{x_0\}) \times S^1$, but it's pretty easy to see that $S^1 \setminus \{x_0\}$ is diffeomorphic to $(0,1)$, which means our set is diffeomorphic to $(0,1) \times S^1$, a cylinder.
I hope this helps ^_^