Suppose $B$ and $C$ are commutative unital $C^{\ast}$-algebras with $B \subseteq C$ (unital). Let $c$ be an element of $C$ such that $c \ast 1 = 1 \ast c$ in the pushout (in the category of commutative unital $C^{\ast}$-algebras) $C \ast\_B C$. Is it valid to conclude that $c$ belongs to $B$?



Elements that commute with $1$ in the pushout of a $C^{\ast}$-algebra
The answer is yes: $1 \ast c = c \ast 1$ implies $c \in B$.




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To see this, let me first translate the statement into fully categorical language. If we restrict to $\|c\| \le 1$ wlog, then the elements $c \in C$ are in bijection with unital $\*$-homomorphisms $\mathscr{C}(D) \to C$, where $D \subseteq \mathbb{C}$ is the unit disk; this is precisely the statement of functional calculus.


Thus, we can formulate the original question as follows: if a morphism $\mathscr{C}(D) \to C$ equalizes the two inclusion maps $C \rightrightarrows C \ast\_B C$, then does itt factor across $B$? If this property even holds for all objects $A$ in place of $\mathscr{C}(D)$, then this is known as the inclusion $B \hookrightarrow C$ being an [effective monomorphism](https://ncatlab.org/nlab/show/effective+monomorphism). It is thus enough to show that this is the case.


Equivalently, under Gelfand duality it is enough to show that every epimorphism in the category of compact Hausdorff spaces is effective. Since this category has pullbacks, it is enough to show that every epimorphism is [regular](https://ncatlab.org/nlab/show/regular+epimorphism). This is well-known to be the case, see e.g. the paper [Algebra $\cap$ Topology = compactness](https://www.sciencedirect.com/science/article/pii/0016660X71900018?via%3Dihub) of Herrlich and Strecker.


