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Let $A$ be a unital $C^\ast$ algebra and let $B$ be a (not necessarily unital) $C^\ast$-subalgebra such that $B \oplus \mathbb C = A$.

I want to argue that the map $\varphi : \widetilde{B} \to A$, $(b,\lambda) \mapsto b + \lambda$ is an isomorphism. It should be obvious but it's not obvious to me.

My thoughts: it is clearly a $\ast$-homomorphism hence I can show that it is isometric from which it follows that it is injective. But given $a\in A$ how can I construct $(b,\lambda)$ such that $b+\lambda =a$?

I also thought about using the isomorphism $B \oplus \mathbb C = A$. Then given $a \in A$, $a$ is identified with some $(b,\lambda)$. The problem is then that nothing guarantees that $b + \lambda = a$.

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  • $\begingroup$ From the background from the previous question, you are following page 44 in Murphy. There $A=B+\mathbb C\,1_A$; it is not a direct sum of C$^*$-algebras (i.e. the product is not coordinate-wise). $\endgroup$ – Martin Argerami Nov 10 '14 at 1:40
  • $\begingroup$ @MartinArgerami I'm sorry could you elaborate? I didn't know there was a difference between $+$ and $\oplus$ in this case (when finitely many terms in the sum). $\endgroup$ – user167889 Nov 10 '14 at 1:52
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Maybe this will clarify things for you. Let $B=C_0(\mathbb R)$. Then you have, as a direct sum of C$^*$-algebras, $$ B\oplus\mathbb C=\left\{\begin{bmatrix}f&0\\ 0&\lambda\end{bmatrix}:\ f\in B,\ \lambda\in\mathbb C\right\}. $$ In that algebra the product is coordinate wise, and the algebra is not unital.

The algebra the Murphy uses in the pages you are using (44-45) is $B+\mathbb C\,1$, i.e. the unitization of $B$, $$ \{f+\lambda:\ f\in B, \ \lambda\in \mathbb C\}\simeq\{(f,\lambda):\ f\in B,\ \lambda\in\mathbb C\}, $$ the latter with the pointwise sum and the product $(f,\lambda)\cdot(g,\mu)=(fg+\lambda g+\mu f,\lambda\mu)$.

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  • $\begingroup$ Thank you for your help! I understand your answer but I'm still confused. If you visit page 12 where unitisation of Banach algebras is defined he explicitly uses $A \oplus \mathbb C$ to denote $\widetilde{A}$. Why can he do that there? $\endgroup$ – user167889 Nov 10 '14 at 2:11
  • $\begingroup$ Because $\oplus$ can mean different things. When you write $B+\mathbb C\,1$ this is a direct sum as vector spaces, but not as C$^*$-algebras. This is very simple to see: in a direct sum of C$^*$-algebras, the product between elements corresponding to different coordinates is zero; this is not the case in the second example above, nor in the unitization. $\endgroup$ – Martin Argerami Nov 10 '14 at 2:13
  • $\begingroup$ I'm sorry for being slow... what are elements corresponding to different coordinates? $\endgroup$ – user167889 Nov 10 '14 at 2:20
  • $\begingroup$ (I understand everything else in your answer and comments, including the answer to my question) $\endgroup$ – user167889 Nov 10 '14 at 3:02
  • $\begingroup$ In the direct sum as algebras, $(a,b)(c,d)=(ac,db)$. So $(a,0)(0,d)=0$. In the sum as in the unitization that cannot be the case, because multiplying by a nonzero scalar can never be zero. $\endgroup$ – Martin Argerami Nov 10 '14 at 4:28

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