8:45 AM
let's say we have charts $U_1, \varphi_1$ and $U_2, \varphi_2$ on our little manifold

9:00 AM
and our transition function $\varphi_2 \circ \varphi_1^{-1}$ is smooth
So we want to build charts $TU_1, \psi_1$ and $TU_2, \psi_2$ on $TM$
$\psi_i(D_{p,v}) := (\varphi_i(p), (D_{p,v}(\varphi^{-1}(t \mapsto \varphi(p)+tx_i)))_{i=1}^n) \in \Bbb R^n \times \Bbb R^n$
is this how it works?
wait that's nonsense
$\psi_i(D_{p,v}) := (\varphi_i(p), (D_{p,v}(u_i \mapsto x^j \cdot \varphi(u_i)))_{j=1}^n) \in \Bbb R^n \times \Bbb R^n$

13 hours later…
9:46 PM
apparently FTC is another special case of Stoke's theorem
$$\int_{\partial M} \omega = \int_M \ \mathrm d\omega$$
ok so let $M=[a,b]$, and we need to put a structure of an oriented manifold with boundary
so we need three charts
I'll actually just do it for $[0,3]$ WLOG
$U_1 = [0,2)$, $U_2 = (1,2)$, $U_3 = (1,3]$
$\varphi_1(x) = x$, $\varphi_2(x) = x$, $\varphi_3(x) = 3-x$
the transition functions are algebraic hence smooth
wait, what's orientation?
this is troublesome
@AlessandroCodenotti @Daminark help me
ok I guess my charts make it into a manifold, and an orientation is a different thing
apparently we only need to orient the interior

10:02 PM
If you have an orientation on a manifold with boundary you should have an induced orientation on the boundary anyway

Cannot right now, I have do some AG :/

so I make a chart $U_4=(0,3)$, $\varphi_4(x)=x$ that orients the interior
anyway we need $\omega \in \Omega^0(M)$, i.e. $f : [0,3] \to \Bbb R$
let $\{\rho_1, \rho_2, \rho_3\}$ be a partition of unity subordinate to $\{U_1, U_2, U_3\}$

10:18 PM
recall $F:M\to N$ induces $F^\ast : T_mM \to T_{F(m)}N$ with $(F^\ast D_v)(g) = D_v(g \circ F)$ and $F_\ast : \Omega^p(N) \to \Omega^p(M)$ with $(F_\ast\omega)_m(D_{v_1}, \cdots, D_{v_p}) = \omega_{F(m)}(F^\ast D_{v_1}, \cdots, F^\ast D_{v_p})$
let's see what $U_3$ becomes
$\displaystyle \int_0^2 ((\varphi_3^{-1})_\ast(\rho_3 \ \mathrm df))(e_1) \ \mathrm dx = \int_0^2 (\rho_3 \ \mathrm df)((\varphi_3^{-1})^\ast e_1) \ \mathrm dx = \int_0^2 (\rho_3 \ \mathrm df)(-e_1) \ \mathrm dx$
then $\displaystyle \int_M \ \mathrm d\omega = \int_0^2 f'(x) \rho_1(x) \ \mathrm dx + \int_1^2 f'(x) \rho_2(x) \ \mathrm dx - \int_0^2 f'(3-x) \rho_3(3-x) \ \mathrm dx = \int_0^3 f'(x) \ \mathrm dx$
modulo some more details that I should have checked
because the LHS is more important

10:44 PM
according to a text I must consider the orientation bundle for the special case of inducing an orientation from dim-1-manifold-with-boundary to dim-0 boundary