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No, and no. The following is a bit too long to write clearly as a comment.

Let $X=\ell^\infty$. Let $c$ be the unital, closed subalgebra of $X$ that consists of all convergent sequences. Let $Y=X/c$ be the quotient algebra and let $A:X\to Y$ be the quotient homomorphism.

Suppose there exists a bounded linear map $B_R:Y\to X$ such that $AB_R$ is the identity map on $Y$. Then $I - B_RA$ would be a bounded linear projection of $\ell^\infty$ onto $c$. But this is known to be impossible by Phillips's lemmaPhillips's lemma.

No, and no. The following is a bit too long to write clearly as a comment.

Let $X=\ell^\infty$. Let $c$ be the unital, closed subalgebra of $X$ that consists of all convergent sequences. Let $Y=X/c$ be the quotient algebra and let $A:X\to Y$ be the quotient homomorphism.

Suppose there exists a bounded linear map $B_R:Y\to X$ such that $AB_R$ is the identity map on $Y$. Then $I - B_RA$ would be a bounded linear projection of $\ell^\infty$ onto $c$. But this is known to be impossible by Phillips's lemma.

No, and no. The following is a bit too long to write clearly as a comment.

Let $X=\ell^\infty$. Let $c$ be the unital, closed subalgebra of $X$ that consists of all convergent sequences. Let $Y=X/c$ be the quotient algebra and let $A:X\to Y$ be the quotient homomorphism.

Suppose there exists a bounded linear map $B_R:Y\to X$ such that $AB_R$ is the identity map on $Y$. Then $I - B_RA$ would be a bounded linear projection of $\ell^\infty$ onto $c$. But this is known to be impossible by Phillips's lemma.

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Yemon Choi
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No, and no. The following is a bit too long to write clearly as a comment.

Let $X=\ell^\infty$. Let $c$ be the unital, closed subalgebra of $X$ that consists of all convergent sequences. Let $Y=X/c$ be the quotient algebra and let $A:X\to Y$ be the quotient homomorphism.

Suppose there exists a bounded linear map $B_R:Y\to X$ such that $AB_R$ is the identity map on $Y$. Then $I - B_RA$ would be a bounded linear projection of $\ell^\infty$ onto $c$. But this is known to be impossible by Phillips's lemma.