Why does the differential $f_{*,p}$ equals $g_{*,p}$ at every point $p\in N$ Hi i am reading An introduction to manifolds by Loring and have 1 doubt in the proof of lemma 9.7. It is written that

the differential $f_{*,p}$ equals $g_{*,p}$ at every point $p\in N$

My question is how do we know that the the differential $f_{*,p}$ equals $g_{*,p}$ at every point $p\in N$? For reference i am attaching the screenshot of proof where i have highlighted the part in which this is mentioned.

Lemma 9.7. Let $g:N\longrightarrow\mathbb{R}$ be a $C^\infty$ function. A regular level set $g^{-1}(c)$ of level $c$ of the function $g$ is the regular zero set $f^{-1}(0)$ of the function $f=g-c$.
Proof. For any $p\in N$,
$$g(p)=c\Longleftrightarrow f(p)=g(p)-c=0.$$
Hence, $g^{-1}(c)=f^{-1}(0)$. Call this set $S$. Because the differential $f_{*,p}$ equals $g_{*,p}$ at every point $p\in N$, the functions $f$ and $g$ have exactly the same critical ponits. Since $g$ has no critical points in $S$, neither does $f$. $\qquad\square$

 A: Note that $f_{\ast,p}$ is a map $T_pN\rightarrow T_{f(p)}\mathbb{R}$, whereas $g_{\ast,p}$ is a map $T_pN\rightarrow T_{g(p)}\mathbb{R}$. So, in a strict sense, these are not equal, because they do not even have the same codomain. In particular, trying to calculate this by definition as you did is doomed to fail. However, there are natural identifications $T_{f(p)}\mathbb{R}\cong\mathbb{R}\cong T_{g(p)}\mathbb{R}$ (which, I'm sure, Tu has introduced before this point) under which the maps in fact become equal.
I believe a conceptual approach clarifies best why this is true. Let $a\colon\mathbb{R}\rightarrow\mathbb{R},\,x\mapsto x-c$ be the auxiliary "subtraction by $c$" map. Then, $f=a\circ g$ by definition, whence $f_{\ast,p}=(a\circ g)_{\ast,p}=a_{\ast,g(p)}\circ g_{\ast,p}$ by the chain rule. However, $a_{\ast,g(p)}$ is an isomorphism, since $a$ is a diffeomorphism. This already implies that $f_{\ast,p}$ is surjective iff $g_{\ast,p}$ is surjective, so that the critical points of $f$ and $g$ are the same. Furthermore, this also lets us see that these maps are the same under the given identifications, since they identify the abstract derivative with the classical derivative from analysis and $a^{\prime}(g(p))=1$.
