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 Yearling
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15m
revised In what kinds of categories is a monic epi an isomorphism?
added argument that implies that a map with a dense image is epi in the category of hausdorff spaces
29m
revised In what kinds of categories is a monic epi an isomorphism?
added 21 characters in body
34m
comment In what kinds of categories is a monic epi an isomorphism?
@ZhenLin - How? Oh! Because if $\phi:X\rightarrow Y$ is not surjective, and $Z$ is an indiscrete space with at least two elements, we can have $\psi,\psi':Y\rightarrow Z$ with $\psi,\psi'$ differing only outside the image of $\phi$, so that $\psi\phi=\psi'\phi$ but $\psi\neq\psi'$. So the idea that a continuous map is determined by its behavior on a dense subset is not true without some separation axiom!
39m
revised In what kinds of categories is a monic epi an isomorphism?
edit to acknowledge Zhen Lin's comment
51m
accepted In what kinds of categories is a monic epi an isomorphism?
1d
comment If $A$ is a finitely generated $R$-module, is $\operatorname{Hom}_R(A,R)$ finitely generated?
@user26857 - [off topic aside] Didn't you post an answer to math.stackexchange.com/questions/378937/… previously? I have been looking forward to the opportunity to think through it carefully but it seems to have been deleted.
1d
comment If $A$ is a finitely generated $R$-module, is $\operatorname{Hom}_R(A,R)$ finitely generated?
@user26857 - I mean that the answer below goes beyond stating the example and also explains the logic behind its construction.
Jul
20
revised If $A$ is a finitely generated $R$-module, is $\operatorname{Hom}_R(A,R)$ finitely generated?
engaged the duplicate issue
Jul
20
accepted If $A$ is a finitely generated $R$-module, is $\operatorname{Hom}_R(A,R)$ finitely generated?
Jul
19
asked If $A$ is a finitely generated $R$-module, is $\operatorname{Hom}_R(A,R)$ finitely generated?
Jul
13
asked In what kinds of categories is a monic epi an isomorphism?
Jul
13
comment Monic and epic implies isomorphism in an abelian category?
@AndreasBlass - I am not following a step in this argument. In a general category, I am with you on this much: if there exists an epic equalizer for two maps, they are equal. But how do you know that it is an isomorphism? In particular, where/how are you deducing the existence of some inverse?
Jul
8
awarded  Yearling
Jul
3
comment Reconciling two different definitions of constructible sets
+1 Thank you! Complete and concise!
Jul
3
accepted Reconciling two different definitions of constructible sets
Jul
1
comment Incorrect notation in math?
Not sure where @columbus8myhw got the $\sqrt[3]{2}$, but I agree with her/him about $2^{1/2^{3/2}}$. By application of $\sqrt[a]{b}=b^{1/a}$, we have $\sqrt[\sqrt{2^3}]{2} = 2^{1/\sqrt{2^3}} = 2^{1/\left(2^{3/2}\right)}=2^{1/2^{3/2}}$.
Jul
1
revised Reconciling two different definitions of constructible sets
Capitalized "Spec"
Jul
1
comment Reconciling two different definitions of constructible sets
@hot_queen - Thanks so much! If you expanded this pair of comments into an answer I would accept it! Also: I am unfamiliar with "the usual clopen basis / usual topology on $2^\omega$" - what does this refer to?
Jun
24
asked Reconciling two different definitions of constructible sets
Jun
19
reviewed Approve Isomorphism between $\Bbb{R}^2 \times \Bbb{R}^2$ and $\Bbb{R}^2 \otimes \Bbb{R}^2$