Universal Property of the Multiplier Algerba

We call an injective $*$-isomorphism $i:A\to B$ between two C*-algebra an *ideal homomorphism* if $i(A)$ is an closed ideal of $B$. The universal property of the multiplier algebra $M(I)$ of a C*-algebra $I$ is: for any C*-algebra $A$ and an ideal homomorphism $i:I\to A$ there is a unique $*$-homomorphism $f$ from $A$ to $M(I)$ such that the following diagram is commutative: \begin{xy} \xymatrix{ A\ar[r]^f & M(I)\\ I\ar[u]^i\ar[ru]_c & , } \end{xy} where $c$ is the canonical ideal homomorphism. *An application*. [If $I$ is a unital C*-algebra, then it cannot be an essential ideal of any other C*-algebra.][1] [1]: https://math.stackexchange.com/q/3046342

left maximal modular ideal in $M_n$

Each pure state on $M_n=B(\mathbb{C}^n)$ is of form $$w_x:=\langle \cdot ~ x, x\rangle,$$ thus each left maximal modular ideal is of form $$\{u|w_x(u^*u)=0\}=\{u|\|ux\|=0\}.$$ Particularly, corresponding to $e_1=(1,0)$ in $\mathbb{C}^2$, the left maximal modular ideal is is of form $$\left( \begin{array} & 0 & a\\ 0 & b \end{array}\right). $$

a Positive Bilinear Map is Necessarily Bounded

Suppose $A$ and $B$ are Banach spaces, $C$ is a normed space, and $\sigma:A\times B\to C$ is a bilinear form. For each $a\in A$, denote the map $b\mapsto \sigma(a,b)$ by $\hat{a}:B\to C$; and for each $b\in B$ denote the map $a\mapsto \sigma(a,b)$ by $\hat{b}:A\to C$. **Lemma**. The bilinear form $\sigma$ is bounded is equivalent to $\hat{a}$ and $\hat{b}$ are both bounded for all $a\in A$ and $b\in B$. *Proof*. Since $\hat{b}$ is a bounded map on $A$ for all elment $b$ in the closed unit ball of $B$, $\sup_{a\in A_1}\|\hat{b}(a)\|\leq \|\hat{b}\|$, i.e., $\sup_{a\in A_1}\|\hat{a}(b)\|\leq \|\hat{b}\|$. By PUB, the family $\{\hat{a}|a\in A_1\}$ of bounded maps is bounded uniformly, that is, $\sup_{a\in A_1}\|\hat a\|\leq M$ for some $M>0$ . Hence, $\sup_{a\in A_1, b\in B_1}\|\sigma(a, b)\|=\sup_{a\in A_1, b\in B_1}\|\hat{a}(b)\|\leq M$. **Application**. Let $A$, $B$ and $C$ be C*-algebras.$\newcommand{\tensor}{\otimes}$ By a *positive* bilinear form $\sigma$ from $A\times B$ to $C$, we mean $\sigma(a,b)\geq 0$ for all positive elements $a$ in $A$ and $b$ in $B$, or equivalently, $\hat{a}$ and $\hat{b}$ are positive linear maps on $B$ and $A$ respectively. Since every positive linear map between two C*-algebra is bounded, by the above lemma, >a positive bilinear form in C*-algebras must be bounded. If $A$ and $B$ are C*-algebras, $\gamma$ is a C*-norm on $A\tensor B$, then the map $A\times B\to A\tensor_\gamma B$ is a positive bilinear form and thus is bounded, that is, $\gamma(a,b)\leq M\|a\|\|b\|$ for some positive number $M$.

the atomic part of $C(X)$

Let $X$ be a compact Hausdorff space and $A=C(X)$. $\newcommand{\id}{\operatorname{id}}\newcommand{\C}{\mathbb{C}}$ For each state $\tau$ of $A$, dnote the associated GNS representation by $(\pi_\tau, H_\tau, x_\tau)$(or $(\pi_\tau,H_\tau)$). If $\tau$ is pure, then $\pi_\tau(A)'=\mathbb{C}\id_\tau$. Since $\pi_\tau(A)\subset \pi_\tau(A)'$, $\pi_\tau(A)=\mathbb{C}\id_\tau$ and thus $\pi_\tau(A)'=B(H_\tau)$. Hence $B(H_\tau)=\C\id_\tau$ and $H_\tau\cong \C$. Suppose $\pi_\tau(a)=k\id_\tau$, then $k=\langle \pi_\tau(a)x_\tau,x_\tau\rangle=\tau(a)$, so $\pi_\tau(a)=\tau(a)\id_\tau$. If $\tau$ and $\rho$ are two pure states of $A$, $(\pi_\tau, H_\tau)$ is unitarily equivalent to $(\pi_\rho, H_\rho)$. Then $\tau(a)\id_\tau=\pi_\tau(a)=u^*\pi_\rho(a) u=\rho(a)u^*\id_\rho u=\rho(a)\id_\tau$,where $u$ is a unitary from $H_\tau$ to $H_\rho$. Therefore, $\tau=\rho$. That is, the atomic representation of $C(X)$ is $(\pi_{atomic}, H_{atomic})=(\oplus_{\tau\in PS(A)}\pi_\tau,\oplus_{\tau\in PS(A)} H_\tau)=(\oplus_{\tau\in \Sigma(A)}\tau(\cdot)\id_\tau, \oplus_{\tau\in \Sigma(A)}\C)=(\oplus_{\tau\in \Sigma(A)}\tau(\cdot)\id_\tau, \ell^2(\Sigma(A))),$ where $\Sigma(A)$ is the character space of $A$. Since $\oplus_{\tau\in PS(A)}\tau(a)\id_\tau=\widehat{a}:\tau\to \tau(a)\id_\tau$, $\pi_{atomic}(A)=C(\Sigma(A))$. As $\pi_{atomic}$ is non-degenerate, operators in $B(\oplus_{\tau\in PS(A)}H_\tau)$ communicated with $\oplus_{\tau\in PS(A)}\pi_\tau(a)$ are of form $\oplus_{\tau\in PS(A)}u_\tau$. By $\|\oplus_{\tau\in PS(A)}u_\tau\|=\sup_{\tau\in PS(A)}\|u_\tau\|< \infty$ we know that $\pi_{atomic}(A)'=\ell^\infty(\Sigma(A))$ and thus $\pi_{atomic}(A)''=\ell^\infty(\Sigma(A))$

An closed left ideal $L$ is the intersection of the maximal modular left ideals containing $L$

**Lemma 1.** Let $L_1$ and $L_2$ be closed ideals of a C* algebra $A$. Suppose that $L_1\subset L_2$ and that every pure state of $A$ that vanishes on $L_1$ vanishes on $L_2$. Then $L_1=L_2$. *Proof.* It is a corollary of Corollary 5.1.9 and Theorem 5.3.2 in [1]. **Lemma 2.** Let $\tau$ be a state of a C*-algebra $A$ and $N_\tau:=\{a|\tau(a^*a)=0\}$. Then a closed left ideal $L$ of $A$ is contained in $N_\tau$ if and only if it is contained in $\ker\tau$. *Proof.* For any element $a$ in $A$, $\tau(a^*a)=0$ implies $\tau(a)=0$ since $|\tau(a)|^2\leq \|\tau\|\tau(a^*a)$. Thus $N_\tau\subset \ker \tau$. Suppose $l\in L$, then $l^*l\in L$ since $L$ is a left ideal. Hence if $L\subset\ker \tau$ then $l^*l \in \ker\tau$, i.e. $l\in N_\tau$. So $L\subset N_\tau$. >**Theorem.** Suppose $L$ is a closed left ideal of a C*-algebra $A$. If the set $$\vee L:=\{N_\tau|\tau \mbox{ is a pure state of } A\}$$ is non-empty, then $L=\cap\vee L$. *Proof.* $L\subset \cap\vee L$ is obvious. Let $\tau$ be a pure state of $A$ and suppose $L\subset \ker\tau$. Since \begin{align*} & L\subset \ker \tau & \\ \Leftrightarrow & L\subset N_\tau & (\mbox{ Lemma 2 })\\ \Rightarrow & \cap\vee L\subset N_\tau & (\mbox{ the definition of } \vee L)\\ \Leftrightarrow & \cap \vee L\subset \ker \tau & (\mbox{ Lemma 2 }), \end{align*} $L=\cap\vee L$ by Lemma 1.


[1] Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.

the bidual of a C*-algebra is linear isometric onto the universal enveloping von Nemann algebra

The following is the collation of the discussion here.

Theorem 1 (3.7.8,[1]). The enveloping von Nemann algebra \(A''\) of a C*-algebra \(A\) is isomorphic, as a Banach space, to the second dual \(A^{\ast \ast}\) of \(A\).

Proof. Let \(S\) denote all the states on the C* algebra and \(\left( {\pi_{S},H_{S}} \right)\) denotes the universal representation of \(A\).

For each state \(\tau \in S\), \(\tau\left( a \right) = \left\langle {\pi_{\tau}\left( a \right)x_{\tau},x_{\tau}} \right\rangle\), where \(\left( {\pi_{\tau},H_{\tau}} \right)\) is the GNS representation associated with \(\tau\) and \(x_{\tau}\) is the cyclic vector for the representation. Hence the vector state \(\left. \tau^{\prime}:u\mapsto\left\langle {ux_{\tau},x_{\tau}} \right\rangle \right.\) in \(B\left( {H\tau} \right)\), which is also normal, makes the triangle above in the following diagram communicative. We know that there exists an normal *-homomorphism from \(B\left( H_{S} \right)\) onto \(B\left( H_{\tau} \right)\), say \(i^{\ast} \circ p\), making the triangle below communicative, Therefore, \(\widetilde{\tau}: = \tau^{\prime} \circ i^{\ast} \circ p\) is an normal extension of \(\tau\) on \(B\left( H_{S} \right)\).

\begin{xy} \xymatrix{ & \mathbb{C}\\ A \ar[ur]^{\tau} \ar[r]_{\pi_\tau} \ar[rd]_{\pi_S}& B(H_\tau) \ar[u]_{{\tau'}} \\ & B(H_S) \ar[u]_{i^*\circ p} } \end{xy}

Since by Jordan Decomposition each element of \(A^{\ast}\) is a linear combination of elements from \(S\) we can therefore define a map \(\Phi\) from \(A^{\ast}\) into \(A_{\ast}'': = \left\{ \tau \in {A{}''}^{\ast} \middle| \tau\text{ is a }\sigma - \text{weakly continous functional on }A'' \right\}\) such that \(\Phi\left( \tau \right) = \widetilde{\tau}|_{A''},\forall\tau \in S\). Note that a state on a von Neumann algebra is normal if and only if it is \(\sigma\)-weakly continuous(3.6.4, [1]).

Since \(\pi_{S}\) is none-degenerate, the strong closure of the closed unit ball of \(A\) is equal to the closed unit ball of \(A''\) by Kaplansky Density Theorem. We also know that the \(\sigma\)-weak topology coincides with the weak topology on a bounded set, and \(WOT\) coincides with \(SOT\) on convex sbuset, hence the \(\sigma\)-weak closure of the closed unit ball of \(A\) is equal to the closed unit ball of \(A''\). Therefore, \(\parallel \widetilde{\tau} \parallel = \parallel \tau \parallel\), i.e., the linear \(\Phi\) is an isometry.

For any \(\tau \in A_{\ast}''\), \(\Phi\left( {\tau|_{A}} \right) = \widetilde{\tau|_{A}}\) coincides with \(\tau\) on \(A\), so \(\Phi\left( {\tau|_{A}} \right) = \tau\). Hence, \(A^{\ast} = A_{\ast}''\). □


[1] Gert Kjaergard. Pedersen. C*-algebras and their automorphism groups. Academic Press London ; New York, 1979.

Closed ideals of $C(X)$ and closed sets of $X$

Theorem 1 (2.13,[1]). Suppose \(V_{1},\cdots\;,V_{n}\) are open subsets of a locally compact Hausdorff space, \(K\) is compact, and

\[K \subset V_{1} \cup \cdots \cup V_{n}.\]

Then there exists functions \(h_{i} \prec V_{i}\left( {i = 1,\cdots\;,n} \right)\) such that \(h_{1}\left( x \right) + \cdots + h_{n}\left( x \right) = 1\left( {x \in K} \right).\)

The collection \(\left\{ {h_{1},\cdots\;,h_{n}} \right\}\) is called a partion of unity on \(K\), subordinate to the cover \(\left\{ {V_{s_{1}},\cdots\;,V_{s_{n}}} \right\}\).

Theorem 2. There is a bijection between closed subsets of \(X\) and closed ideals of \(C\left( X \right)\), where \(X\) is a compact Hausdorff space.

Proof. Suppose \(I\) is a closed ideal of \(C\left( X \right)\). Let \(I^{\top}: = \left\{ x \in X \middle| f\left( x \right) = 0\forall f \in I \right\}\), and \(\varepsilon > 0\). If \(s \notin I^{\top}\), then there is a \(f_{s}^{\prime} \in I\) such that \(f_{s}^{\prime}\left( s \right) \neq 0\) and thus there is a \(f \in I\) such that \(f_{s}\left( s \right) = 1\). Let \(V_{s} = \left\{ x \middle| \middle| f_{s}\left( x \right) - 1 \middle| < \varepsilon \right\}\), then \(V_{s},s \in K\) is a open cover of \(K\). Hence we have a finite open cover \(V_{s_{1}},\cdots\;,V_{s_{n}}\) of \(K\).

Let \(\left\{ {h_{1},\cdots\;,h_{n}} \right\}\) is called a partion of unity on \(K\) subordinate to the cover \(\left\{ {V_{s_{1}},\cdots\;,V_{s_{n}}} \right\}\), and let

\[f_{K}: = \sum\limits_{i = 1}^{n}f_{s_{i}}h_{i}.\]

Then \(f_{K} \in I\), and \(\left| f_{K}\left( x \right) - 1 \middle| = \middle| \sum_{i = 1}^{n}f_{s_{i}}\left( x \right)h_{i}\left( x \right) - \sum_{i = 1}^{n}h_{i}\left( x \right) \middle| \leq \sum_{i = 1}^{n} \middle| f_{s_{i}}\left( x \right) - 1 \middle| h_{i}\left( x \right) \leq \varepsilon\left( {x \in K} \right) \right.\).

For any \(g \in I^{\top\bot}: = \left\{ f \in C\left( X \right) \middle| f\left( x \right) = 0\forall x \in I^{\top} \right\}\), set \(K = \left\{ x \in X \middle| \middle| g\left( x \right) \middle| \geq 0 \right\}\). Since

\[\begin{array}{rlrl} \left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \right| & \left. \leq \parallel g \parallel \middle| 1 - f_{K}\left( x \right) \middle| < \parallel g \parallel \varepsilon\left( {x \in K} \right),\qquad \right. & & \qquad \\ \left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \right| & \left. \leq \middle| g\left( x \right) \middle| \left( 1 + \middle| f_{s_{i}}\left( x \right) \middle| h_{i}\left( x \right) \right)\qquad \right. & & \qquad \\ & {\leq \varepsilon\left( \left. 1 + \middle| f_{s_{i}}\left( x \right) \right| \right) \leq \varepsilon\left( {2 + \varepsilon} \right)\left( {x \in \cup_{i = 1}^{n}V_{s_{i}} \backslash K} \right),\qquad} & & \qquad \\ \left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \right| & \left. = \middle| g\left( x \right) \middle| < \varepsilon\left( {x \notin \cup_{i = 1}^{n}V_{s_{i}}} \right),\qquad \right. & & \qquad \\ \end{array}\]

\(\left| g\left( x \right) - g\left( x \right)f_{K}\left( x \right) \middle| \leq \left( {\parallel g \parallel + 2 + \varepsilon} \right)\varepsilon \right.\). Therefore, \(g \in I\) and \(I^{\top\bot} = I\).

Suppose \(B\) is a closed set in \(X\). If there exists some \(x_{0} \in B^{\bot\top} \backslash B\), then by Uryshon’s lemma, there exists some \(f \in C\left( X \right)\) such that \(f\left( x \right) = 0\forall x \in B\) and \(f\left( x_{0} \right) \neq 0\), i.e., \(f \in B^{\bot}\) but \(x_{0} \notin B^{\bot\top},\) a contradiction. Hence \(B^{\bot\top} = B\). □


[1]   Walter Rudin. Real and complex analysis. Tata McGraw-Hill Education, 1987.

Jordan Decomposition of bounded linear functionals on a C*-algebra

Let’s first review Theorem 3.3.6,[1].

Theorem 1. If \(a\) is a normal element of a non C*-algebra \(A\), then there exists a state \(\tau\) of \(A\) such that \(\left| \tau\left( a \right) \middle| = \parallel a \parallel \right.\).

Corollary 2. Suppose \(A\) is a C*-algebra and \(a \in A\). If \(\tau\left( a \right) = 0\) for all state \(\tau\) of \(A\), then \(a = 0\).

Proof. Suppose \(a = b + ic\) where \(b\) and \(c\) are hermitian elements of \(A\). For any state \(\tau\) on \(A\), we have \(\tau\left( b \right),\tau\left( c \right) \in {\mathbb{R}}\), thus if \(\tau\left( a \right) = \tau\left( b \right) + i\tau\left( c \right) = 0\), then \(\tau\left( b \right) = 0,\tau\left( c \right) = 0\). By the arbitrariness of \(\tau\) and Theorem 1, \(b = c = 0\). □

By an involutive vector space, we mean a complex vector space \(V\) with an involution, that is, a conjugate-linear operator \(\left. \ast :V\rightarrow V \right.\) such that \(\ast \circ \ast = \operatorname{Id}\). An element \(v \in V\) is called self-adjoint if \(v^{\ast} = v\).

Lemma 3. Suppose \(\left. T:V\rightarrow W \right.\) is a homomorphism between two involutive vector spaces \(V,W\). If for a self-adjoint element \(w \in W\), there exists some \(v \in V\) such that \(T\left( v \right) = w\), then there is a self-adjoint element \(v^{\prime} \in V\) such that \(T\left( v^{\prime} \right) = w\).

Proof. Let \(v^{\prime}: = \frac{v + v^{\ast}}{2}\) and \(v^{''} = \frac{v - v^{\ast}}{2i}\) be self-adjoint elements in \(V\), then \(v = v^{\prime} + iv^{''}\). Since \(w = T\left( v \right) = T\left( v^{\prime} \right) + iT\left( v^{''} \right)\) is self-adjoint, \(T\left( v^{''} \right) = 0\), thus \(w = T\left( v \right) = T\left( v^{\prime} \right)\). □

Theorem 4 (Jordan decomposition). Let \(\tau\) be a self-adjoint bounded linear functional on a C*-algebra, then there exist some positive linear functionals \(\tau_{+},\tau_{-}\) such that \(\tau = \tau_{+} - \tau_{-}\) and \(\parallel \tau \parallel = \parallel \tau_{+} \parallel + \parallel \tau_{-} \parallel\).

Proof. Let \(S\) be the set of all the states on \(A\), and

\[\begin{matrix} {\Phi:} & \left. A\rightarrow C\left( S \right) \right. & & \\ & \left. a\mapsto\widehat{a}: \right. & \left. S\rightarrow{\mathbb{C}} \right. & \\ & & \left. \tau\mapsto\tau\left( a \right). \right. & \\ \end{matrix}\]

then \(\Phi\) has following properties:

(a) \(\Phi\) is an injection. \(\left. \Phi\left( a \right) = \Phi\left( b \right)\Rightarrow\tau\left( a \right) = \tau\left( b \right) \right.\) for all states. By Corollary 2, a=b.

(b) \(\Phi\) preserves involution: \(\tau\left( a^{\ast} \right) = \overline{\tau\left( a \right)}\) for all states and \(\widehat{a^{\ast}}\left( \tau \right) = \tau\left( a^{\ast} \right)\), \({\widehat{a}}^{\ast}\left( \tau \right) = \overline{\widehat{a}\left( \tau \right)} = \overline{\tau\left( a \right)}.\)

(c) \(\Phi\) is linear and bounded.

(d) \(\Phi\) is bounded below. Suppose \(a \in A\) and \(a = b + ic\), where \(b = \frac{a + a^{\ast}}{2}\) and \(c = \frac{a - a^{\ast}}{2i}\) are hermitian element of \(A\). Then

\[\begin{array}{rlrl} & {\left. \parallel \widehat{a} \parallel = \sup\limits_{\tau \in S} \middle| \tau\left( a \right) \middle| = \sup\limits_{\tau \in S} \middle| \tau\left( {b + ic} \right) \right|\qquad} & & \qquad \\ = & {\sup\limits_{\tau \in S}\tau\left( b \right) + i\tau\left( c \right) \geq \sup\limits_{\tau \in S}\max\left\{ \left| \tau\left( b \right) \middle| , \middle| \tau\left( c \right) \right| \right\}\qquad} & & \qquad \\ = & {\max\left\{ {\parallel b \parallel , \parallel c \parallel} \right\} \geq \frac{1}{2} \parallel b + ic \parallel \geq \frac{1}{2} \parallel a \parallel .\qquad} & & \qquad \\ \end{array}\]


\[\begin{array}{rlrl} {\Phi^{\ast}:} & \left. C\left( S \right)^{\ast}\rightarrow A^{\ast}\qquad \right. & & \qquad \\ & \left. \mu\mapsto\mu \circ \Phi\qquad \right. & & \qquad \\ \end{array}\]

then \(\Phi^{\ast}\) has the following properties: (a) \(\Phi^{\ast}\) is linear.

(b) \(\Phi^{\ast}\) is preserves involution. \(\Phi^{\ast}\left( \mu^{\ast} \right)\left( a \right) = \mu^{\ast}\left( {\Phi\left( a \right)} \right) = \overline{\mu\left( \left( {\Phi\left( a \right)} \right)^{\ast} \right)} = \overline{\mu\left( {\Phi\left( a^{\ast} \right)} \right)},\) and \(\left( {\Phi^{\ast}\left( \mu \right)} \right)^{\ast}\left( a \right) = \overline{\Phi^{\ast}\left( \mu \right)\left( a^{\ast} \right)} = \overline{\mu\left( {\Phi\left( a^{\ast} \right)} \right)}\)\(\left( {\mu \in C\left( X \right)^{\ast},a \in A} \right).\)

(c) \(\Phi^{\ast}\) is a surjection. Since \(\Phi\) is s bounded below, \(\left. \Phi^{- 1}:\Phi\left( A \right)\rightarrow A \right.\) is bouded. Suppose \(\tau \in A^{\ast}\), then \(\tau \circ \Phi^{- 1} \in \Phi\left( A \right)^{\ast}\). Let \(\mu \in A^{\ast}\) be the extension of \(\tau \circ \Phi^{- 1}\), then \(\Phi^{\ast}\left( \mu \right) = \mu \circ \Phi = \tau \circ \Phi^{- 1} \circ \Phi = \tau\).

For a self-adjoint bounded linear functional \(\tau \in A^{\ast}\), by lemma 3, there exists some self-ajoint linear functional \(\mu \in C\left( S \right)^{\ast}\) such that \(\Phi^{\ast}\left( \mu \right) = \tau\). It corresponds to a real regular measure on \(S\). □


[1]   Gerald J Murphy. C*-algebras and operator theory. Academic press, 2014.

A compact Hausdorff space $X$ is homeomorphic to the maximal ideal space of $C(X)$

$\newcommand{\M}{\operatorname{Max}}\newcommand{\C}{\mathbb{C}}$**Theorem**. If $X$ is a compact Hausdorff space, and $\M=\{\text{non-zero algebra homomorphism of }C(X) \text{to } \C\}$ is the maximal ideal space of $C(X)$, then there is a homeomorphism of $X$ onto $\M$. *Proof*. For each $x\in X$, define \begin{equation*} ev_x(f)=f(x), f\in C(X), \end{equation*} we have Claim. \begin{equation*} H:X\to \M, x\to ev_x \end{equation*} is a homeomorphism. (a)Obviously, $H$ is well-defined. (b)By Urysohn's lemma, $H$ is an injection. (c)If $\{x_\lambda\}$ is a net in $X$ converges to $x\in X$, then $f(x_\lambda)\to f(x)$ for all $f\in C(X)$, i.e., $H(x_\lambda)\xrightarrow{weak*}H(x)$. Hence $H$ is continuous. (d)Let $I$ be a none-trivial ideal of $C(X)$ and denote all the zero points of $f\in C(X)$ by $Z(f)$. For any finite continuous functions $f_i\in I, i=1,2,\cdots, n$, $$f:=\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n\in I,$$ therefore $ f$ is not invertible, and thus $Z(f)\neq\emptyset$. Since $$Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\supset Z(\bar{f_1}f_1+\bar{f_2}f_2+\cdots \bar{f_n}f_n),$$ $Z(f_1)\cap Z(f_2)\cap\cdots\cap Z(f_n)\neq\emptyset.$ Hence we have $$\bigcap_{f\in I}Z(f)\neq\emptyset.$$ Note that $Z(f)$ is closed for every continuous function and that $X$ is compact. In particular, take $I=\ker\tau, \tau\in \M$. Suppose $x\in \bigcap_{f\in I}Z(f)$, then $\ker\tau=I\subset \{f\in C(X)|f(x)=0\}=\ker ev_x$, so $\tau=ev_x$. Therefore, $H$ is a surjection. (e) Since $X$ is compact and $\M$ is Hausdorff, $H$ is a homeomorphism.