[proofplan] We define the candidate extension by the only possible formula: send the adjoined unit of $A^+$ to the unit of $B$ and agree with $\pi$ on $A$. The main verification is algebraic: using the unitization multiplication and involution formulas, the candidate preserves products, adjoints, scalar combinations, and the unit. Uniqueness follows from the decomposition $(a,\lambda)=(a,0)+\lambda(0,1)$, and the injectivity criterion follows by solving the equation $\pi(a)+\lambda 1_B=0$. [/proofplan]

[step:Define the only possible unital extension] Define a map $\pi^+:A^+\to B$ by \begin{align*} \pi^+(a,\lambda)=\pi(a)+\lambda 1_B \end{align*} for every $a\in A$ and every $\lambda\in\mathbb C$. This map extends $\pi$ because, for every $a\in A$, \begin{align*} \pi^+(a,0)=\pi(a). \end{align*} It is unital because the unit of $A^+$ is $(0,1)$, and hence \begin{align*} \pi^+(0,1)=1_B. \end{align*} [/step]

[step:Verify that the extension is a unital $*$-homomorphism]Let $(a,\lambda),(b,\mu)\in A^+$ and let $\alpha,\beta\in\mathbb C$. Since $\pi:A\to B$ is linear, we have \begin{align*} \pi^+(\alpha(a,\lambda)+\beta(b,\mu))=\alpha\pi^+(a,\lambda)+\beta\pi^+(b,\mu). \end{align*} Using the unitization product and the multiplicativity and linearity of $\pi$, we compute \begin{align*} \pi^+((a,\lambda)(b,\mu))=\pi(ab+\lambda b+\mu a)+\lambda\mu 1_B. \end{align*} Therefore \begin{align*} \pi^+((a,\lambda)(b,\mu))=\pi(a)\pi(b)+\lambda\pi(b)+\mu\pi(a)+\lambda\mu 1_B. \end{align*} Since $1_B$ is the multiplicative identity in $B$, the right-hand side is \begin{align*} (\pi(a)+\lambda 1_B)(\pi(b)+\mu 1_B)=\pi^+(a,\lambda)\pi^+(b,\mu). \end{align*} Thus $\pi^+$ preserves multiplication. Using the unitization involution and the fact that $\pi$ preserves adjoints, we also have \begin{align*} \pi^+((a,\lambda)^*)=\pi(a^*)+\overline{\lambda}1_B. \end{align*} Hence \begin{align*} \pi^+((a,\lambda)^*)=\pi(a)^*+\overline{\lambda}1_B=(\pi(a)+\lambda 1_B)^*=\pi^+(a,\lambda)^*. \end{align*} Together with $\pi^+(0,1)=1_B$, this proves that $\pi^+:A^+\to B$ is a unital $*$-homomorphism.[/step]

[guided]The only point that needs checking is that the formula respects the algebra structure of the unitization. Let $(a,\lambda),(b,\mu)\in A^+$ and let $\alpha,\beta\in\mathbb C$. The [vector space](/page/Vector%20Space) structure on $A^+=A\oplus\mathbb C$ is componentwise, so \begin{align*} \alpha(a,\lambda)+\beta(b,\mu)=(\alpha a+\beta b,\alpha\lambda+\beta\mu). \end{align*} Using linearity of $\pi:A\to B$, we obtain \begin{align*} \pi^+(\alpha(a,\lambda)+\beta(b,\mu))=\pi(\alpha a+\beta b)+(\alpha\lambda+\beta\mu)1_B. \end{align*} This equals \begin{align*} \alpha(\pi(a)+\lambda1_B)+\beta(\pi(b)+\mu1_B)=\alpha\pi^+(a,\lambda)+\beta\pi^+(b,\mu). \end{align*} Thus $\pi^+$ is linear. Now we check multiplication. The product in the standard unitization is \begin{align*} (a,\lambda)(b,\mu)=(ab+\lambda b+\mu a,\lambda\mu). \end{align*} Applying $\pi^+$ gives \begin{align*} \pi^+((a,\lambda)(b,\mu))=\pi(ab+\lambda b+\mu a)+\lambda\mu1_B. \end{align*} Because $\pi$ is a $*$-homomorphism, it is linear and multiplicative. Therefore this becomes \begin{align*} \pi(a)\pi(b)+\lambda\pi(b)+\mu\pi(a)+\lambda\mu1_B. \end{align*} On the other hand, multiplying the two proposed images in the unital algebra $B$ gives \begin{align*} \pi^+(a,\lambda)\pi^+(b,\mu)=(\pi(a)+\lambda1_B)(\pi(b)+\mu1_B). \end{align*} Expanding in $B$ and using that $1_B$ is the identity gives the same expression: \begin{align*} \pi^+(a,\lambda)\pi^+(b,\mu)=\pi(a)\pi(b)+\lambda\pi(b)+\mu\pi(a)+\lambda\mu1_B. \end{align*} Hence $\pi^+$ preserves multiplication. Finally, the involution in $A^+$ is \begin{align*} (a,\lambda)^*=(a^*,\overline{\lambda}). \end{align*} Since $\pi$ preserves adjoints, \begin{align*} \pi^+((a,\lambda)^*)=\pi(a^*)+\overline{\lambda}1_B=\pi(a)^*+\overline{\lambda}1_B. \end{align*} Because $1_B^*=1_B$, this equals \begin{align*} (\pi(a)+\lambda1_B)^*=\pi^+(a,\lambda)^*. \end{align*} Together with $\pi^+(0,1)=1_B$, this proves that $\pi^+$ is a unital $*$-homomorphism.[/guided]

[step:Prove uniqueness from the decomposition in the unitization] Let $\rho:A^+\to B$ be a unital $*$-homomorphism such that $\rho(a,0)=\pi(a)$ for every $a\in A$. For every $(a,\lambda)\in A^+$, the vector space decomposition in $A^+=A\oplus\mathbb C$ gives \begin{align*} (a,\lambda)=(a,0)+\lambda(0,1). \end{align*} Using linearity of $\rho$ and unitality of $\rho$, we obtain \begin{align*} \rho(a,\lambda)=\rho(a,0)+\lambda\rho(0,1)=\pi(a)+\lambda1_B. \end{align*} Thus $\rho=\pi^+$, so the unital extension is unique. [/step]

[step:Characterize when the extension is injective] Assume that $\pi$ is injective. First suppose that $\pi(A)$ contains no nonzero scalar multiple of $1_B$. Let $(a,\lambda)\in A^+$ satisfy $\pi^+(a,\lambda)=0$. Then \begin{align*} \pi(a)+\lambda1_B=0. \end{align*} If $\lambda=0$, then $\pi(a)=0$, and injectivity of $\pi$ gives $a=0$. Hence $(a,\lambda)=(0,0)$. If $\lambda\ne 0$, then \begin{align*} \pi(a)=-\lambda1_B. \end{align*} This says that $\pi(A)$ contains the nonzero scalar multiple $-\lambda1_B$ of $1_B$, contradicting the hypothesis. Therefore the case $\lambda\ne 0$ cannot occur, and $\ker\pi^+=\{(0,0)\}$. Hence $\pi^+$ is injective. Conversely, suppose that $\pi(A)$ contains a nonzero scalar multiple of $1_B$. Then there exist $a\in A$ and $\lambda\in\mathbb C\setminus\{0\}$ such that \begin{align*} \pi(a)=\lambda1_B. \end{align*} For the element $(a,-\lambda)\in A^+$, we have \begin{align*} \pi^+(a,-\lambda)=\pi(a)-\lambda1_B=0. \end{align*} Also $(a,-\lambda)\ne(0,0)$ because its scalar component is $-\lambda\ne 0$. Thus $\pi^+$ has a nonzero kernel element and is not injective. Combining both directions, $\pi^+$ is injective if and only if $\pi(A)$ contains no nonzero scalar multiple of $1_B$. [/step]