How is the Algebra $C_{p}^{\infty}(U)$ of germs of $C^\infty$ functions in $U$ at $p$ is the Same as $C_{p}^{\infty}(M)$ Hi i am reading An introduction to manifolds by Loring and have some doubts in remark 8.2. It is written that

If $U$ is an open set containing $p$ in $M$ then the algebra $C_{p}^{\infty}(U)$ of germs of $C^\infty$ functions in $U$ at $p$ is the same as $C_{p}^{\infty}(M)$.Hence, $T_pU=T_pM$.

My first question is:

*

*How do we know that the algebra $C_{p}^{\infty}(U)=C_{p}^{\infty}(M)$. I know that, the equivalence class of $(f,U)$ is called the germ of $f$ at $p$.

I have a second question which is related to the differential of a map. It is written that:

The equation $(F_{*}(X_p))f=X_p(f\circ F)$ is independent of the representative of the germ.

My second question is that


*here in the above equation $f$ is the representative of a germ and it is appearing inside the equation then how this equation is independent of the representative $f$?

For reference i am attaching the screenshots where i have highlighted the part where these two statements are mentioned.

 A: For $X= M, U$ let $G_p^\infty(X)$ denote the set of
all pairs $(f ,V)$, where $U$ is an open neighborhood of $p$ in $X$ and $f : V \to \mathbb R$ is a $C^\infty$-function. Then $C_p^\infty(X) = G_p^\infty(X)/\sim_X \phantom{.}$ where $(f,V) \sim_X (g,W)$ if there exists an open neighborhood $W$ of $p$ in $X$ such that $W \subset U \cap V$ and  $f \mid_W = g \mid_W$.
Since $U$ is open in $M$, we have $G_p^\infty(U) \subset G_p^\infty(M)$, but if $U \subsetneqq M$, then certainly $G_p^\infty(M)$ has elements $(V,f)$ such that $V \not\subset U$, i.e. $G_p^\infty(M)$ has more elements than $G_p^\infty(U)$. For $(V,f), (W,g) \in G_p^\infty(U)$ we have $(V,f) \sim_U (W,g)$ if and only if $(V,f) \sim_M (W,g)$, i.e. we get an injective function
$$i : C_p^\infty(U) \to C_p^\infty(M), i([(V,f)]_U) = [(V,f)]_M $$
which is easily seen to be an algebra homomorphism. This map is in fact a bijection: Given $(V,f) \in G_p^\infty(M)$, we have $(V,f) \sim_M (V \cap U, f \mid_{V \cap U})$. But clearly $(V \cap U, f \mid_{V \cap U}) \in G_p^\infty(U)$.
More formally, let $\phi : N \to M$ be smooth. Then $\phi$ induces a function $G^\infty_p \phi : G^\infty_{f(q)} M \to G^\infty_q N, G^\infty_p \phi (U,f) = (\phi^{-1}(U), f \circ \phi\mid_{\phi^{-1}(U)})$. Clearly $(U,f) \sim_M (V,g)$, then $G^\infty_p \phi (U,f) \sim_N G^\infty_p \phi (V,g)$. Therefore we get an induced
$$C^\infty_p \phi : C^\infty_{f(q)} M \to C^\infty_q N$$
which is easily seen to be an algebra homomorphism.
Now let $\iota :  U \to M$ denote the inclusion map which is clearly smooth. Then $C^\infty_p \iota : C^\infty_p M \to C^\infty_p U$ is easily seen to be the inverse algebra isomorphism of $i : C^\infty_p U \to C^\infty_p M$.
In that sense $C_p^\infty(U)$ is the same as $C_p^\infty(M)$. But be aware that, strictly speaking, we do not have $C_p^\infty(U) = C_p^\infty(M)$, instead we have the canonical algebra isomorphism $i : C_p^\infty(U) \to C_p^\infty(M)$ with inverse $C^\infty_p \iota : C^\infty_p M \to C^\infty_p U$.
Similarly it is an abuse of notation to write $T_pU = T_pM$, but again we have a canonical  isomorphism $T_pU \to T_pM$.
In fact, each algebra homomorphism (isomorphism) $\mathfrak h :  \mathfrak A \to \mathfrak B$ induces a vector space homomorphism (isomorphism)
$$\mathscr D(\mathfrak h) :  \mathscr D(\mathfrak B) \to \mathscr D(\mathfrak A), \mathscr D(\mathfrak h)(D)  = D \circ \mathfrak h .$$
If we take our above canonical algebra isomorphism $i : C_p^\infty(U) \to C_p^\infty(M)$ we get a canonical vector space isomorphism $\mathscr D(i) : T_p M =  \mathscr D(C_p^\infty(M)) \to T_pU = \mathscr D(C_p^\infty(U))$ with inverse
$$\mathscr D(C^\infty_p \iota) : T_pU \to T_pM .$$
The isomorphism $\mathscr D(C^\infty_p \iota)$ is nothing else than the differential $d_p\iota$ of $\iota$ at $p$. So writing $T_pU = T_pM$ is a sloppy way to say that $d_p\iota : T_pU \to T_pM$ is a canonical isomorphism.
Concerning your second question:
$X_p$ is a derivation on $C_p^\infty(N)$. Given $(V,f) \in G_{F(p)}^\infty(M)$, the map $f \circ F : F^{-1}(V) \to \mathbb R$ gives us an element $(F^{-1}(V),f \circ F) \in G_p^\infty(N)$. It is easy to see that if $(V,f) \sim_M (W,g)$, then $(F^{-1}(V),f \circ F) \sim_N (F^{-1}(W),g \circ F)$. This means that the map $C_{F(p)}^\infty(M) \to C_p^\infty(N), [(V,f)]_M \mapsto [(F^{-1}(V),f \circ F)]_N $ is well-defined.
Instead of $(F_{*}(X_p))f=X_p(f\circ F)$ Tu should have written
$$(F_{*}(X_p))([(V,f)]_M) = X_p([(F^{-1}(V),f\circ F)]_N) .$$
But I think his little abuse of notation is acceptable.
Update:
The above algebra isomorphism $i : C_p^\infty(U) \to C_p^\infty(M)$ has the special feature that its definition is completely obvious and natural. In fact, it does not involve any choices, but is induced by the set inclusion $G_p^\infty(U) \hookrightarrow G_p^\infty(M)$. If you consider germ algebras $C_p^\infty(M), C_q^\infty(N)$ for manifolds $M, N$ of dimension $n
$, you will see that they are always isomorphic, but in general all isomorphisms $\beta: C_p^\infty(M) \to C_q^\infty(N)$ involve choices which do not make $\beta$ "natural" in any sense. In fact, you can choose charts $\phi : U \to B^n$ around $p$ on $M$ and $\psi : V \to B^n$ around $q$ on $N$ (where $B^n$ is an open ball in $\mathbb R^n$ with center $0$) such that $\phi(p) = \psi(q) = 0$. Then $C_p^\infty(M) \approx C_p^\infty(U)$ and  $C_q^\infty(N) \approx C_q^\infty(V)$ in a natural way and $C_p^\infty(U) \approx C_0^\infty(B^n), C_q^\infty(V) \approx C_0^\infty(B^n)$, but the last two isomorphisms are induced by $\phi, \psi$ (which have be chosen arbitarily), i.e. are not natural in  our sense.
For that reason one frequently writes $C_p^\infty(U) = C_p^\infty(M)$ which would be inadequate in general for $C_p^\infty(M)$ and $C_q^\infty(N)$.
This transfers to tangent spaces: $T_pU$ and $T_pM$ are isomorphic in a unique obvious way, the vector space isomorphism being induced by the algebra isomorphism $i: C_p^\infty(U) \to C_p^\infty(M)$ and its inverse $C^\infty_p \iota : C_p^\infty(M) \to C_p^\infty(U)$. For that reason one frequently writes $T_pU  = T_pM$. Note that all tangent spaces $T_pM$ and $T_qN$ are isomorphic, but finding an isomorphism involves choices which prevents writing $T_pM = T_qN$.
