C¹ germs of local diffeomorphisms form a group

(Well-definedness of composition.) Let $[f]=[f^{\mathrm{\prime }}]$ and $[g]=[g^{\mathrm{\prime }}]$ be germs at $x$, with $f,g$ fixing $x$. Choose neighbourhoods $A,B$ of $x$ on which $f=f^{\mathrm{\prime }}$ and $g=g^{\mathrm{\prime }}$, respectively. Since $g(x)=g^{\mathrm{\prime }}(x)=x$ and both maps are continuous, choose an open neighbourhood $W\subseteq B$ of $x$ with $g(W)\cup g^{\mathrm{\prime }}(W)\subseteq A$; then for $y\in W$, $(f\circ g)(y)=f(g(y))=f^{\mathrm{\prime }}(g^{\mathrm{\prime }}(y))=(f^{\mathrm{\prime }}\circ g^{\mathrm{\prime }})(y)$, so $f\circ g{\sim }_x{f}^{\mathrm{\prime }}\circ g^{\mathrm{\prime }}$. Hence the operation on germs is well defined; it is associative because composition of maps is associative.

(Identity and inverses.) The germ of ${\mathrm{i}\mathrm{d}}_T$ at $x$ is a two-sided identity for the operation. If $f:U\to f(U)$ is a representative, then $f^{-1}:f(U)\to U$ is again a $C^1$ local diffeomorphism fixing $x$ [F2], and its germ depends only on the germ of $f$: if $f^{\mathrm{\prime }}$ agrees with $f$ on a neighbourhood $W\subseteq U\cap U^{\mathrm{\prime }}$ of $x$, then $f^{-1}$ and $f^{\mathrm{\prime }-1}$ agree on the open set $f(W)\cap f^{\mathrm{\prime }}(W)$, which contains $x$. Thus every germ has the two-sided inverse given by the class of any representative's inverse, and ${\mathrm{Diff}}_x^1(T)$ is a group [F4]. The Euclidean inverse function theorem identifies the same local inverses in a chart at $x$ [F3].

(The positive-derivative germs.) Fix an oriented chart $t$ at $x$ with $t(x)=0$ and write $\tilde{f}$ for the coordinate expression of a representative. The sign of ${\tilde{f}}^{\mathrm{\prime }}(0)$ is independent of the positively oriented chart and of the representative, since a positive change of coordinate $\psi$ contributes ${\psi }^{\mathrm{\prime }}(0)>0$ and its inverse likewise, so it does not change the sign [F1]. By the chain rule, $(f\circ g){}^{\mathrm{\prime }}(0)=f^{\mathrm{\prime }}(0)g^{\mathrm{\prime }}(0)>0$ and $(f^{-1}{)}^{\mathrm{\prime }}(0)=1\mathrm{/}{f}^{\mathrm{\prime }}(0)>0$ whenever $f^{\mathrm{\prime }}(0)>0$ and $g^{\mathrm{\prime }}(0)>0$ [F5], and the identity has derivative $1$. Hence the germs of positive derivative contain the identity and are closed under composition and inverses, so by the subgroup criterion they form a subgroup ${\mathrm{Diff}}_x^{1,+}(T)\le {\mathrm{Diff}}_x^1(T)$ [F4].

Composition is a well-defined associative operation with identity and inverses, so ${\mathrm{Diff}}_x^1(T)$ is a group, and the positive-derivative germs form the subgroup ${\mathrm{Diff}}_x^{1,+}(T)$.