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visits member for 2 years, 7 months
seen Jan 18 at 3:23

Jan
17
answered Non-compact manifold with compact boundary
Jan
17
comment Proving that S/I is a vector space
$I + I$ is not the same as $2I$. In the first case you get $x + y$ for any $x$, $y$; in the second case you only get $x + x$ for any $x$. So $2I$ is smaller in general, and is not relevant to the proof. You should be able to show that $I + I \subseteq I$ easily by looking carefully at what it means for $I$ to be an ideal.
Jan
17
accepted What are the ring homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$?
Jan
11
revised What are the ring homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$?
clarified definitions
Jan
11
comment What are the ring homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$?
@Lubin, yes, that's correct, I mean the set of all functions $\mathbb N \to \mathbb Z$ with componentwise operations.
Jan
11
asked What are the ring homomorphisms $\mathbb{Z}^\mathbb{N} \to \mathbb{Z}$?
Jan
11
comment nicer proof of basic functor category fact?
Also, I notice that in the books I'm using, this exercise appears just before Yoneda's Lemma, which suggests that we should be able to do the exercise without using Yoneda. But I'm curious about any approach which works.
Jan
11
comment nicer proof of basic functor category fact?
@QiaochuYuan, can you say more about what you mean by $\operatorname{Hom}(X, Y^Z) \cong \operatorname{Hom}(X \times Z, Y)$? Is this an isomorphism between functors $\operatorname{Cat}^3 \to \operatorname{Set}$, or maybe just a bijection between sets?
Jan
2
comment nicer proof of basic functor category fact?
Thanks, I'll look into that. How would you remove the size issues in general? In order to use this approach, must we assume that $\mathcal A$, $\mathcal B$, $\mathcal C$ are objects in some larger category?
Jan
1
comment Finding $6$ pairwise non-isomorphic semi-direct products of order $42$
Oh, and don't forget that the action can be trivial, so ${\Bbb Z}_{42}$ and other abelian groups should probably be on your list.
Jan
1
comment Finding $6$ pairwise non-isomorphic semi-direct products of order $42$
The second one you're looking for seems to be the dihedral group of size 42, which you get by letting ${\Bbb Z}_2$ act on ${\Bbb Z}_{21}$ by inverting it.
Jan
1
answered $x^n y^n = (xy)^n$, proof exercise
Jan
1
asked nicer proof of basic functor category fact?
Dec
29
revised name this “hybrid” categorical construction
clarified
Dec
29
accepted name this “hybrid” categorical construction
Dec
29
answered name this “hybrid” categorical construction
Dec
23
comment Inverse image of a functor
This does look similar to the image of a morphism, but presumably in a functor category.
Dec
15
comment Is the suspension space contractible?
Consider upvoting or accepting my answer if it's helpful!
Dec
13
revised Inverse image of a functor
edited body
Dec
13
asked Inverse image of a functor