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1d

reviewed  Leave Open Is a group uniquely determined by the sets $\{ab,ba\}$ for each pair of elements a and b? 
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answered  Are there examples of families of objects which are canonically isomorphic, but where diagrams of canonical isomorphisms don't commute? 
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reviewed  Leave Open A metric associated with a continuous surjective map $f:X\to Y$ 
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answered  A metric associated with a continuous surjective map $f:X\to Y$ 
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awarded  Enlightened 
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awarded  Nice Answer 
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comment 
How did the summation operation come into use?
What do you mean by "the summation operation" exactly? 
Oct 21 
reviewed  Close About freeness of modules over the coordinate ring of an affine variety 
Oct 19 
answered  Is every polynomial ring over any field regular? 
Oct 18 
comment 
“Nice” functions on infinitedimensional space of germs of continuous functions at a point
Can you even construct any linear functional on germs of continuous functions (besides multiples of evaluation at the point) without the axiom of choice? 
Oct 14 
comment 
Connected graph as connected space
The closure of any connected set is connected. 
Oct 12 
comment 
Connected graph as connected space
Any compactification of a connected space is connected. 
Oct 9 
comment 
Classification of rings satisfying $a^4=a$
@MartinBrandenburg: It is a Stone space if you topologize it based on $\mathbb{F}_4$ being discrete, rather than only $\{0\}$ being closed. 
Oct 7 
awarded  Nice Answer 
Oct 6 
awarded  Yearling 
Oct 1 
answered  Standard homology result on double complexes 
Sep 29 
comment 
What is the most useful nonexisting object of your field?
@AndréHenriques: Actually, it has $2^8$ elements. But you are correct that free complete Boolean algebras on finite sets exist (and are the same as free Boolean algebras). Free complete Boolean algebras on infinite sets do not exist. 
Sep 29 
reviewed  Looks OK What is the most useful nonexisting object of your field? 
Sep 23 
comment 
A simple example of a ring that is an $A$module but not an $A$algebra?
I'm not sure why you say $\mathbb{R}$ is a $\mathbb{C}$vector space but not a $\mathbb{C}$algebra; any embedding of $\mathbb{C}$ in $\mathbb{R}$ makes $\mathbb{R}$ a $\mathbb{C}$algebra. In your situation, $B$ will be an $A$algebra iff the action of $A$ on $B$ commutes with the action of $B$ on itself. Equivalently, $B$ is an $A$algebra iff the action of $a\in A$ coincides with multiplication by $a\cdot 1_B$ in the ring structure of $B$. 
Sep 9 
answered  Completion of the set of subsets with half volume. 