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comment About equalizer of Boolean algebras
No, they aren't the same in general. For instance, let $I$ be any connected compact Hausdorff space, and let $\pi:S\to I$ be any continuous surjection from a Stone space to $I$. Taking the kernel pair of $\pi$ gives a pair of maps $\varphi^*,\psi^*:T\to S$ from some Stone space $T$ whose coequalizer is $\pi$. Thus the quotient by the equivalence relation $\sim'$ in this case is $I$. But the quotient by $\sim$ will be just a point, since $I$ is connected.
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answered About equalizer of Boolean algebras
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answered Description of $\big(\ell^\infty(\mathbb N)\big)^{\!*}$ via ultrafilters
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comment $K$-Theory of finite dimensional Banach algebras
If you allow non-unital algebras, it is easy to find a finite-dimensional algebra $A$ such that $K_0(A)$ is trivial (for instance, let $A=\mathbb{C}$ with the zero multiplication; then over $\tilde{A}=\mathbb{C}[x]/(x^2)$ all projective modules are free). Whether $K_0(A)$ can be finite but nontrivial seems a lot harder.
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revised Partial Orders realized by Prime Ideals on commutative rings
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revised Partial Orders realized by Prime Ideals on commutative rings
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revised Partial Orders realized by Prime Ideals on commutative rings
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answered Partial Orders realized by Prime Ideals on commutative rings
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revised Cantor's theorem for presheaves?
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revised Who gave the generalized Stone-Weierstrass Theorem?
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revised Who gave the generalized Stone-Weierstrass Theorem?
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answered Who gave the generalized Stone-Weierstrass Theorem?
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