In order to reduce notational confusion, the composition of two mappings $f$ and $g$ is denoted $f \bullet g$ instead of the more usual $f \circ g$. There are two cases to consider: when $T_1$ and $S_2$ are disjoint, and when they are not. Suppose $T_1$ and $S_2$ are disjoint. Let $S_1$ be the set $T_1 \cup \left({S_2 \setminus T_2}\right)$. Then we can define the mapping $h: S_2 \to S_1$ as: :$\forall x \in S_2: h \left({x}\right) = \begin{cases} x & : x \in S_2 \setminus T_2 \\ f^{-1} \left({x}\right) & : x \in T_2 \end{cases}$ Because $T_1$ and $S_2$ are disjoint, $h$ can be seen to be a bijection. What $h$ is doing is that it effectively "slots" $T_1$ into the gap in $S_2$ that was taken up by the removed $T_2$. We are going to show that $\left({T_1, \oplus_1}\right)$ is embedded in $\left({S_1, *_1}\right)$. The operation $*_1$ is defined as the transplant of $*_2$ under $h$. Thus by the Transplanting Theorem: :$\forall x, y \in S_1: x *_1 y = h \left({h^{-1} \left({x}\right) *_2 h^{-1} \left({y}\right)}\right)$ Let $x, y \in T_1$. Then, from the definition of $h$ above: :$h^{-1} \left({x}\right) = \left({f^{-1} \left({x}\right)}\right)^{-1} = f \left({x}\right)$ and similarly: :$h^{-1} \left({y}\right) = f \left({y}\right)$ So: {{begin-eqn}} {{eqn | l = x *_1 y       | r = h \left({f \left({x}\right) *_2 f \left({y}\right)}\right)       | c = Transplanting Theorem }} {{eqn | r = h \left({f \left({x}\right) \oplus_2 f \left({y}\right)}\right)       | c = as $*_2$ is embedded in $\oplus_2$ }} {{eqn | r = h \left({f \left({x \oplus_1 y}\right)}\right)       | c = Morphism Property of $\oplus_1$ under $f$ }} {{eqn | r = f^{-1} \left({f \left({x \oplus_1 y}\right)}\right)       | c = Definition of $h$ on elements of $T_1$ }} {{eqn | r = x \oplus_1 y       | c = Inverse of Inverse of Bijection }} {{end-eqn}} proving that $x \oplus_1 y$ is closed. Thus: : $\left({T_1, \oplus_1}\right) \subseteq \left({S_1, *_1}\right)$ that is, $\left({T_1, \oplus_1}\right)$ is embedded in $\left({S_1, *_1}\right)$. Now, let $g = h^{-1}$. By the definition of $*_1$, $g$ is an isomorphism from $\left({S_1, *_1}\right)$ onto $\left({S_2, *_2}\right)$. Then: {{begin-eqn}} {{eqn | l = x \in T_1       | o = \implies       | r = f \left({x}\right) \in T_2       | c = by definition of $f$ }} {{eqn | o = \implies       | r = h \left({f \left({x}\right)}\right) = f^{-1} \left({f \left({x}\right)}\right) = x       | c = by definition of $h$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = g \left({x}\right)       | r = g \left({h \left({f \left({x}\right)}\right)}\right)       | c = from above }} {{eqn | r = h^{-1} \left({h \left({f \left({x}\right)}\right)}\right)       | c = by definition of $g$ }} {{eqn | r = f \left({x}\right)       | c = Inverse of Inverse of Bijection }} {{end-eqn}} So $g$ is an extension of $f$. We have therefore proved that the embedding theorem holds when $T_1$ and $S_2$ are disjoint. {{qed|lemma}} Next, suppose $T_1$ and $S_2$ are ''not'' disjoint. We can use Exists Bijection to a Disjoint Set to assume the existence of a bijection $k: S_2 \to S_3$ such that $S_3 \cap T = \varnothing$. Let $*'$ be the transplant of $*$ under $k$. Let: :$T_3 = \left\{{k \left({x}\right): x \in T_2}\right\}$ Then: {{begin-eqn}} {{eqn | l = x, y \in T_3       | o = \implies       | r = k^{-1} \left({x}\right), k^{-1} \left({y}\right) \in T_2       | c = by definition of $k$ }} {{eqn | o = \implies       | r = k^{-1} \left({x}\right) * k^{-1} \left({y}\right) \in T_2       | c = by definition of $*$ in $T_2$ }} {{eqn | o = \implies       | r = x *' y = k \left({k^{-1} \left({x}\right) * k^{-1} \left({y}\right)}\right) \in T_3       | c = Transplanting Theorem }} {{end-eqn}} So $T_3$ is closed under $*'$. Let $*' {\restriction_{T_3} }$ be the operation on $T_3$ induced by $*'$. Then $\left({T_3, *' {\restriction_{T_3} } }\right)$ is embedded in $\left({S_3, *'}\right)$. Thus $k \circ f$ is an isomorphism from $\left({T_1, \oplus_1}\right)$ onto $\left({T_3, *' {\restriction_{T_3} } }\right)$. $S_3$ has been constructed so as to be disjoint from $T_1$. In such a case it has already been shown that the embedding theorem holds. That is: :$(1): \quad$ there exists a magma $\left({S_1, *_1}\right)$ containing $\left({T_1, \oplus_1}\right)$ algebraically :$(2): \quad$ there exists an isomorphism $g_1$ from $\left({S_1, *_1}\right)$ to $\left({S_3, \oplus_1}\right)$ extending $k \bullet f$. Let $g = k^{-1} \bullet g_1$. Let $x \in T$. Then: {{begin-eqn}} {{eqn | l = g \left({x}\right)       | r = k^{-1} \left({g_1 \left({x}\right)}\right)       | c = definition of $g$ }} {{eqn | r = k^{-1} \left({k \left({f \left({x}\right)}\right)}\right)       | c = as $g_1$ extends $k \bullet f$ }} {{eqn | r = f \left({x}\right)       | c = Inverse of Inverse of Bijection }} {{end-eqn}} and so $g$ extends $f$. As $k$ is an isomorphism from $\left({S_2, *}\right)$ onto $\left({S_3, *'}\right)$, then: :$(1): \quad g$ is an isomorphism from $\left({S_1, *_1}\right)$ onto $\left({S_2, *}\right)$ :$(2): \quad g$ extends $f$. Hence the result. {{qed}} {{Proofread}}