Remark 2

The characteristic functions $\varphi_M$ and $\varphi_N$ of two subsets $M$ and $N$ of $X$ have the same germ with respect to $\mathfrak{F}$ if and only if $M$ and $N$ have the same germ with respect to $\mathfrak{F}$.

Let $X''$ be a third set, $\varphi$ a mapping of $X'$ into $X''$, $\Phi'$ the set of all mappings of a set of $\mathfrak{F}$ into $X''$. For each $f \in \Phi$, $\varphi \circ f$ belongs to $\Phi'$; further, it is immediately seen that if $g \in \Phi$ has the same germ as $f$ with respect to $\mathfrak{F}$, then $\varphi \circ f$ and $\varphi \circ g$ have the same germ with respect to $\mathfrak{F}$.

This germ therefore depends only on the germ $\tilde{f}$ of $f$ with respect to $\mathfrak{F}$ and is denoted by $\varphi(\tilde{f})$. We thus define a mapping (denoted by $\varphi$, by abuse of language) of the set $\tilde{\Phi}$ of germs of mappings of $X$ into $X'$, into the set $\tilde{\Phi}'$ of germs of mappings of $X$ into $X''$.

Now let $X'_i$ ($1 \leq i \leq n$) be sets and

their product; let $\Phi_i$ (resp. $\Phi$) denote the set of all mappings of a set of $\mathfrak{F}$ into $X'_i$ (resp. $Y$). If $f_i \in \Phi_i$ for $1 \leq i \leq n$ and if $M_i \in \mathfrak{F}$ is the domain and $f_i$, then the mapping $t \to (f_1(t), \ldots, f_n(t))$ is defined on and hence belongs to $\Phi$; we denote this mapping (by abuse of language) by $(f_1, \ldots, f_n)$. Furthermore, if $f_i$ and $g_i$ belong to $\Phi_i$ and have the same germ with respect to $\mathfrak{F}$ (for $1 \leq i \leq n$), it is immediately seen that $(f_1, \ldots, f_n)$ and $(g_1, \ldots, g_n)$ have the same germ with respect to $\mathfrak{F}$; this germ therefore depends only on the germs $\tilde{f}_i$ of the $f_i$. If we denote it by $\Gamma(\tilde{f}_1, \ldots, \tilde{f}_n)$ then $\Gamma$ is clearly a bijection of the product set onto the set $\tilde{\Phi}$, where $\tilde{\Phi}_i$ (resp. $\tilde{\Phi}$) denotes the set of germs of mappings of $X$ into $X'_i$ (resp. $Y$) with respect to $\mathfrak{F}$; hence, by abuse of language, we shall generally write $(\tilde{f}_1, \ldots, \tilde{f}_n)$ instead of $\Gamma(\tilde{f}_1, \ldots, \tilde{f}_n)$ whenever there is no risk of confusion.

From what has been said, every mapping $\psi$ of $Y$ into a set $X''$ defines a mapping $(\tilde{f}_1, \ldots, \tilde{f}_n) \to \psi(\tilde{f}_1, \ldots, \tilde{f}_n)$ of

into the set $\tilde{\Phi}'$ of germs of mappings of $X$ into $X''$.

In particular, if $n = 2$ and if $X'_1, X'_2$ and $X''$ are all equal to the same set $X'$ (so that $\psi$ is a law of composition defined throughout $X'$), then $\psi$ induces a law of composition defined throughout the set $\tilde{\Phi}$ of germs of mappings of $X$ into $X'$. It is easily verified that if the law given on $X'$ is associative (resp. commutative) then so is the corresponding law on $\tilde{\Phi}$; if the law $\psi$ on $X'$ has an identity element $e'$, then the germ with respect to $\mathfrak{F}$ of the constant mapping $x \to e'$ is an identity element for the corresponding law on $\tilde{\Phi}$. Finally, if the law on $X'$ has an identity element $e'$, then the germ $\tilde{f}$ of $f \in \Phi$ has an inverse in $\tilde{\Phi}$ if and only if there exists $V \in \mathfrak{F}$, contained in the domain of $f$, such that $f(t)$ is invertible in $X'$ for each $t \in V$; if, for each $t \in V$, $g(t)$ denotes the inverse of $f(t)$ then the germ $\tilde{g}$ of $g$ is the inverse of $\tilde{f}$ in $\tilde{\Phi}$. In particular if $X'$ is a group with respect to the law $\psi$, then $\tilde{\Phi}$ is a group with respect to the corresponding law; likewise, if $X'$ is a ring (resp. an algebra over a ring $A$) then $\tilde{\Phi}$ is a ring (resp. an algebra over $A$) with respect to the corresponding laws of composition.