Differentiable manifolds

Two charts $(U,x),(V,y)$ are $\text{D}$ compatible if either:

• $U\cap Y=\mathrm{\varnothing }$
  or:
• $y\circ x^{-1}{|}_{U\cap V}$ and $x\circ y^{-1}{|}_{U\cap V}$ are both differentiable under the same notion $\text{D}$

In this case, The atlas with all such charts may be denoted ${\mathcal{A}}_D$. The triple $(M,{\mathrm{\Theta }}_M,{\mathcal{A}}_D)$ is then a differentiable manifold under the notion $D$.

If there $C^k$- manifold $(M,\mathrm{\Theta },A_{C^k})$ for $k\ge 1$, then there is a sub-atlas of $A_{C^k}$, which is a $C^{\mathrm{\infty }}$ atlas.

If $(M,{\mathrm{\Theta }}_M,A_M),(N,{\mathrm{\Theta }}_N,A_N)$ are both smooth manifolds, of dimensions $m,n$. Then $\varphi :M\to N$ is a $C^{\mathrm{\infty }}$ map if for any chart $(U,x)\in A_M$ and any chart $(V,y)\in A_N$, The charted map $g_{U,V}:{\mathbb{R}}^m\supset x(U)\to y(V)\subset {\mathbb{R}}^n$ is a smooth function.
Notice that $\varphi {|}_{U\to V}=y^{-1}\circ g\circ x$. Or $g=y\circ \varphi {|}_{U\to V}\circ x^{-1}$.

Moreover, if there are alternative charting maps $\tilde{x}$ on $U$ and $\tilde{y}$ on $V$, then it better be that $h=\tilde{y}\circ \varphi {|}_{U\to V}\circ {\tilde{x}}^{-1}$ is also a smooth map. This holds good, because the chart transition maps between $x,\tilde{x}$ and $y,\tilde{y}$ are smooth maps.
More formally, $\tilde{y}=t_{y\to \tilde{y}}\circ y$ and ${\tilde{x}}^{-1}=x^{-1}\circ t_{x^{-1}\to {\tilde{x}}^{-1}}$. Hence, $h=t_{y\to \tilde{y}}\circ g\circ t_{x^{-1}\to {\tilde{x}}^{-1}}$. But since both of the transition maps is smooth, and $g$ is smooth, $h$ is also smooth.

If $(M,{\mathrm{\Theta }}_M,A_M),(N,{\mathrm{\Theta }}_N,A_N)$ are both smooth manifolds, then they are diffeomorphic if there exists a bijection $\varphi :M\to N$ such that both $\varphi$ and ${\varphi }^{-1}$ are $C^{\mathrm{\infty }}$ maps. It is a very cool theorem that given a $C^0$ manifold, only for dimension $4$ we have an uncountably infinite number of sub-atlases that gives us a smooth manifold (upto diffeomorphism) from the given $C^0$ manifold.

Let $(M,{\mathrm{\Theta }}_M,A_M)$ be a smooth manifold, $\mathbb{R}$ is trivially a smooth manifold (with the identity map $id:\mathbb{R}\to \mathbb{R}$ serving as the single global chart, and hence no chart transition maps to consider).
Then the set of all smooth maps ($C^{\mathrm{\infty }}$) maps from $M\to \mathbb{R}$ is called $C^{\mathrm{\infty }}(M)$, and can be equipped with a vector space structure. With the usual function addition and scalar multiplication we all know and love.

For each $p\in M$ of a smooth manifold, define the set $T_{p}M=\{v_{\gamma ,p}:\text{for all smooth curves}\gamma \}$

So you have all these different smooth curves, each giving tangent vectors at the point $p$ on the smooth manifold. So we can get addition and scalar multiplication on $T_{p}M$ easily
In general the result is a linear map from $C^{\mathrm{\infty }}(M)$ to $\mathbb{R}$, but we want the result to be another vector in the tangent space, namely we want the result in $T_{p}M$ itself, to get closure on the vector space.
for any $\lambda \in \mathbb{R}$ and any tangent vector $v_{\gamma ,p}$ there exists a curve $\delta$ such that for all $f\in C^{\mathrm{\infty }}(M)$, $v_{\delta ,p}(f)=\lambda v_{\gamma ,p}(f)$ That is, $$ (f\circ\delta)'(y_{0}) = \lambda(f \circ \gamma)'(x_{0})$$ Where $\delta (y_0)=\gamma (x_0)=p\in M$.
What about introducing a map $\mathbb{R}\stackrel{h}{⟶}\mathbb{R}\stackrel{\gamma }{⟶}M\stackrel{f}{⟶}\mathbb{R}$
such that $(f\circ \gamma \circ h{)}^{\prime }(y_0)=\lambda (f\circ \gamma {)}^{\prime }(x_0)$? By the chain rule, the equation simplifies to $$ (f\circ \gamma)'(h(y_{0}))h'(y_{0}) = \lambda(f\circ \gamma)'(x_{0})$$
(Here $\delta =\gamma \circ h$, where since $\gamma$ is smooth, and $h$ will chosen to be smooth, $\delta$ will be smooth too)
But what $h$ do we pick? what about $h(t)=\lambda t+x_0$? firstly let us solve for $y_0$, we want $\gamma (h(y_0))=\gamma (x_0)$ it is sufficent if $h(y_0)=x_0$, and with our choice for $h$, $y_0=0$. In any case $h^{\prime }(y_0)=\lambda$ and $h(0=y_0)=x_0$, giving the desired inequality.

Now comes the harder part:
pick any two elements of $T_{p}M$ say $v_{\gamma ,p},v_{\delta ,p}$. We must show that there exists a smooth curve $h$ such that for all "temperature" $C^{\mathrm{\infty }(M)}$/smooth maps from $f:M\to \mathbb{R}$ $$v_{\gamma,p}(f) + v_{\delta,p}(f) = v_{h,p}(f)$$.
Its time for A CRAZY setup again. First, we will make the curve $h$ chart dependent, and then see if we can correct that sin.
Pick a chart $(U,x)$ such that $p\in U$. choose $t_0,t_1\in \mathbb{R}$ such that $\gamma (t_0)=\delta (t_1)=p$
now define: $$ h_{x}(t) = x^{-1}\left((x \circ \gamma)(t_{0} + t) + (x \circ \delta)(t_{1}+t)- (x \circ \gamma)(t_{0})\right )$$
Let $h_x$ reminds us that our definition of the curve $h$ is still chart dependent. It is clear that indeed $h_x(0)=p$ *by the way, the whole curves $\gamma ,\delta$ need not lie in the chart open set $U$, because only the point $p$ and thus only $h_x(0)$ is what we want to compute on. *

So looking at the commutative diagram,$h_x$ uses maps from $\mathbb{R}\to {\mathbb{R}}^d$ namely $(x\circ \gamma ),(x\circ \delta )$ and then inverts to the open set $U$ in the manifold. Now, $$ v_{h_{x},p} (f) = (f \circ h_{x})'(0)$$as usual, we introduce the identity to use undergrad notion of derivatives in vector spaces: $x^{-1}\circ x$ giving us

Looking at the commutative diagram above again, $(f\circ x^{-1})=p:{\mathbb{R}}^d\to \mathbb{R}$ and $(x\circ h_x)=q:\mathbb{R}\to {\mathbb{R}}^d$. So both these parts are "undergaduate derivative compatible".
Now we shut up and calculate.
if we peturb an input to $q$, a $d$ dimensional vector is perturbed as the output, which is then fed into $p$. Let us look at a computational graph.