What is “Free”

**Update**.Suppose $\mathbf{B}$ is a category, $\mathbf{A}$ is a subcategory of $\mathbf{B}$, and $i:X\to G$ is an injective morphism. If for each object $Y$ of $\mathbf{A}$ and each morphism $f:X\to Y$ there is a unique morphism $\tilde{f}:G\to Y$ such that the following diagram is commutative: \begin{xy} \xymatrix{ X\ar[r]^f \ar[d]_i&Y\\ G\ar[ur]_{\tilde{f}} & , } \end{xy} then $G$ is said to be a *free* object with a *basis* $X$. ------ Suppose $\mathbf{A}$ is a category, $G$ is an object of $\mathbf{A}$. If there is a parent category $\mathbf{B}$ of $\mathbf{A}$ with an object $X$ and an injective morphism $i:X\to G$ satisfying that for any object $Y$ of $\mathbf{A}$ and each morphism $f:X\to Y$ there is a unique morphism $\tilde{f}:G\to Y$ such that the following diagram is commutative: \begin{xy} \xymatrix{ X\ar[r]^f \ar[d]_i&Y\\ G\ar[ur]_{\tilde{f}} & , } \end{xy} then $G$ is said to be a *free* object. (discussed with S.Q.Huang)

the category of open projections and that of hereditary C*-subalgebras


$\CI$: - Obejects: hereditary C*-subalgebras $\CI(A)$ of a C*-algebra $A$. - Arrows: Subset orthomorphisms. By a *subset orthomorphism* between $\CI(A)$ and $\CI(B)$, we mean a mapping $\Theta$ that preserrves orthogonality, in the sense $IJ=0$ implies that $\Theta(I)\Theta(J)=0$ for all $I, J\in \CI(A)$.

**The Functor between the Two Categories**: We know that the mapping $F:p\mapsto pA''p\cap A$ from $\OP(A)$ onto $\CI(A)$ is a bijection, where $A''$ is the universal enveloping von Neumannn algebra of $A$. The functor from $\OP$ to $\CI$ is defined to be a function $\F$ assigns to each projection orthomorphism $\theta:\OP(A)\to \OP(B)$ a subset orthomorphism $$\F\theta:\CI(A)\to \CI(B), F(p)\mapsto F(\theta(p)).$$ It is a isomorphism between the two categories. We must prove that for each subset isomorphism $\Theta:\CI(A)\to \CI(B)$, $\theta: \CI(A)\to \CI(B), F^{-1}(H)\mapsto F^{-1}(\Theta(H))$ is a projection orthomorphism. Suppose $F(p), F(q)\in \CI(A)$ satisfies that $F(p)F(q)=0$. Then any approximate unit $(u_\lambda)$ of $F(p)$ will strongly converges to the unit $p$ of $pA''p,$ the strong closure of $F(p)$. Since the composition of operators in a bounded set is continuous, $pq=0$. \begin{xy} \xymatrix{ \OP(A)\ar[r]^{F_A}\ar[d]_\theta & \CI(A)\ar[d]^{\F\theta}\\ \OP(B)\ar[r]^{F_B} & \CI(B) } \end{xy}