Tensor product of fixed ring isomorphism. Question :
Let $A$ be a ring and $\phi : A\to A$ be a ring homomorphism that is an involution. Suppose $A$ is a finitely generated $\mathbb{C}$ algebra under the map $\beta : \mathbb{C} \to A$ and 
$\psi: \mathbb{C} \to  \mathbb{C}$ be conjugation map such that $\phi\circ\beta=\beta\circ\psi$.
Define $B=[a\in A : \phi(a)=a]$. The map $\beta$ then restricts to $\mathbb{R}\to B$. Then show that $B\otimes_\mathbb{R}\mathbb{C}$ is isomorphic to $A$.
Attempt :
Define the map $\Delta : B\otimes_\mathbb{R}\mathbb{C}\to A$ as sending an element $b\otimes z$ to $b\beta(z)$. 
I am trying to show that this map is injective and surjective but I am not able to prove either of them. 
For example, for injectivity, if $b\beta(z)=0$, then how can I conclude that $b\otimes z=0$.
 A: Note that $A$ is a $\Bbb C$ vector space with scalar multiplication defined as $$z \cdot a = \beta(z)a$$
for $z \in \Bbb C$ and $a \in A$. This makes $A$ an $\Bbb R$ vector space as well.
Since $\phi$ is a homomorphism, it is clear that $B$ is closed under addition (and that it contains $0$). The fact that $\beta$ restricts as a map from $\Bbb R$ to $B$ shows that $B$ is actually an $\Bbb R$-subspace of $A$. (Indeed, if $x \in \Bbb R$ and $b \in B$, then $$\phi(x \cdot b) = \phi(\beta(x)b) = \phi(\beta(x))\phi(b) = \beta(\psi(x))b = \beta(x)b = x \cdot b$$ and thus, $x \cdot b \in B$.)
Thus, showing that $B \otimes_{\Bbb R} \Bbb C \cong A$ is showing the isomorphism of two $\Bbb R$ vector spaces. This can be done by showing that the dimensions are equal. Thus, we need to show that $$2\dim_{\Bbb R}B = \dim_{\Bbb R}A.$$
(Since $\dim_{\Bbb R}\Bbb C = 2$.)
Define $$\iota B = \{\iota \cdot b \mid b \in B\}.$$
Clearly, the above is an $\Bbb R$-subspace of $A$. Moreover, $\dim_{\Bbb R}B = \dim_{\Bbb R}(\iota B)$.
Claim 1. $B \cap \iota B = 0$.
Proof. Let $x \in B \cap \iota B$. In particular, $x = \iota \cdot y$ for some $y \in B$. Now, note that
\begin{align}
x = \phi(x) &= \phi(\iota\cdot y)\\
&= \phi(\beta(\iota)y)\\
&= \phi(\beta(\iota))\phi(y)\\
&= \beta(\psi(\iota)) y\\
&= \beta(-\iota)y\\
&= (-\iota)\cdot y\\
&= -(\iota\cdot y) = -x.
\end{align}
Thus, $x = 0$. $\blacksquare$
Claim 2. If $a \in A$ and $\phi(a) = -a$, then $a \in \iota B$.
Proof. It suffices to show that $\iota^{-1}\cdot a = -\iota \cdot a \in B$.
To this end, note that
\begin{align}
\phi(-\iota \cdot a) &= \phi(\beta(-\iota) a)\\
&= \phi(\beta(-\iota))\phi(a)\\
&= \beta(\psi(-\iota))(-a)\\
&= \beta(\iota)(-a)\\
&= -\iota\cdot a.
\end{align}
Thus, $-\iota\cdot a \in B$. $\blacksquare$
Claim 3. $A = B \oplus \iota B$.
Proof. By Claim 1, we only to show that $+$ part. To this end, note that every $a \in A$ can be written as
$$a = \underbrace{\dfrac{1}{2}\cdot(a + \phi(a))}_{\in B} + \underbrace{\dfrac{1}{2}\cdot(a - \phi(a))}_{\in \iota B}.$$
To see the underbraced statements, simply apply $\phi$ to the term and use the fact that $\phi(\phi(a)) = a$ ($\phi$ is an involution). This directly gives that the first term is indeed in $B$. For the second term, note that
$$\phi\left(\dfrac{1}{2}\cdot(a - \phi(a))\right) = -\dfrac{1}{2}\cdot(a - \phi(a))$$
and thus, $\dfrac{1}{2}\cdot(a - \phi(a)) \in \iota B$, by Claim 2. $\blacksquare$
Thus, it follows that
$$\dim_{\Bbb R}A = \dim_{\Bbb R}B + \dim_{\Bbb R}\iota B = 2\dim_{\Bbb R}B,$$
as desired.
