Let $A$ be a unital $C^{\ast}$-algebra and $\{ f_i: A \rightarrow A_i \}_i$ a finite collection of morphisms of unital $C^{\ast}$-algebras, such that the associated map $A \rightarrow \prod_i A_i$ is injective. Let $g: A \rightarrow B$ be a morphism of unital $C^{\ast}$-algebras. Then, there is an induced collection of maps $\{ B \rightarrow A_i \ast_A B \}_i$, where $A_i \ast_A B$ is the amalgamated free product (pushout). This collection induces a map $f: B \rightarrow \prod_i A_i \ast_A B$. Is $f$ injective?
$\textbf{Motivation:}$ I am trying to use the extensive Grothendieck topology in the category of unital $C^{\ast}$-algebras.
Best Answer
Unfortunately, the assertion is, in general, not true even if we just have a single (injective) unital $*$-homomorphism $f_1\colon A\to A_1$ and another (non-injective) unital $*$-homomorphism $g\colon A\to B$. The reason is that amalgamated free products with non-injective maps can be trivial: $B*_A A_1=0$ in certain cases.
For an example, take any simple unital $C^*$-algebra $A_1$, for example $A_1=M_2(\mathbb{C})$ the $2\times 2$-matrices, and $A$ any unital $C^*$-subalgebra of $B$ with $\dim(A)>1$ and which has a character $\chi\colon A\to \mathbb{C}$, for example the diagonal matrices $D\cong \mathbb{C}^2$ in $M_2(\mathbb{C})$.
Now take $B=\mathbb{C}$ with the character $g=\chi\colon A\to B$ above, and take $f_1$ to be the embedding $A\subseteq A_1$.
In this situation, the amalgamated free product
$$A_1*_A B$$
is canonically isomorphic to the quotient $A_1/J$, where $J$ is the ideal of $A_1$ generated by the differences $\chi(a)1_{A_1}-a$ with $a\in A$. Since $\dim(A)>1$, this ideal is not zero, so that $J=A_1$ because $A_1$ is simple and therefore
$$A_1*_A B\cong A_1/J =0.$$
In particular, for the specific example mentioned above:
$$M_2(\mathbb{C})*_{\mathbb{C}^2}\mathbb{C}=0.$$