Let $\mathcal{C}^\star$ denote the category that has $C^*$-algebras as objects and $*$-homomorphisms as morphisms. My question is the following: Are monomorphisms in $\mathcal{C}^\star$ with the categorical sense precisely injective $*$-homomorphisms?
Obviously, if $A\xrightarrow{\psi}B$ is an injective $*$-homomorphism, then $\psi$ is a monomorphism. Indeed, if $C\xrightarrow{h_1}A$, $C\xrightarrow{h_2}A$ are $*$-homomorphisms such that $f\circ h_1=f\circ h_2$, then $f(h_1(x))=f(h_2(x))$ for all $x$, so, since $f$ is injective we get $h_1(x)=h_2(x)$ for all $x$, so $h_1=h_2$.
Conversely, suppose that $A\xrightarrow{f}B$ is a mono. Let $x,x'\in A$ with $f(x)=f(x')$ and suppose that $x\neq x'$. I consider $C:=C^*(x,x')$, the $C^*$-subalgebra of $A$ generated by $x$ and $x'$. We want to consider two $*$-homomorphisms $C\to A$ such that their compositions with $f$ are equal. The one would simply be the inclusion $C\xrightarrow{inc}A$. I was thinking that maybe I could define the other one as a flip-map, i.e. maybe I could define $h:C\to A$ starting by defining $h$ on the $*$-subalgebra generated by $x,x'$ by setting $h(x)=x'$ and $h(x')=x$ and then extending as a $*$-homomorphism to the entire $*$-subalgebra that is generated by $x,x'$. If I could show that $h$ is bounded there, then $h$ would extend to a $*$-homomorphism $C\to A$. Also since $h\circ f=inc\circ f$ on the set of generators $\{x,x'\}$, this would yield equality on all $C$. This would give me a contradiction and this would all prove that one-to-one $*$-homomorphisms are precisely the monomorphisms of $\mathcal{C}^\star$.
However, I feel that my idea won't work. Does anybody know if the result is even true? If so, am I on the right path to prove it? If not, could anybody provide a proof?
Regards
A simplification to your idea is to note that you need to prove that $f(x)=0$ implies $x=0$. Now you can take $g_1$ the inclusion map and $g_2=0$. You obtain automatically that $x=0$.