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Jun
29
comment Singular $\simeq$ Cellular homology?
@MartinBrandenburg One still has to prove that there is indeed a quasi-isomorphism...
Jun
29
comment Singular $\simeq$ Cellular homology?
Quasi-isomorphic implies homotopy equivalent in this situation, because both chain complexes are degreewise projective.
Jun
29
comment Polynomial Functor is ω-Continuous
Again, I'm not sure what you're trying to do. There is a canonical $X_i \times X_i \to X \times X$, by functoriality. If you really want to do things by abstract nonsense, show that the diagonal is cofinal (in the sense of category theory) in $\omega \times \omega$. Of course, you will have to give separate arguments for $\omega \times \omega \times \omega$ and so on.
Jun
28
comment Polynomial Functor is ω-Continuous
Yes, $\omega$ is filtered. You have to do some work here, and indeed, some playing around with elements and equivalence relations and so on. There is a whole section in [CWM] about filtered colimits – you should read it.
Jun
28
comment Polynomial Functor is ω-Continuous
I don't know what you're trying to do. There's no point trying to prove that $X \mapsto X \times X$ preserves colimits – it doesn't: $1 \times 1 + 1 \times 1 \ne (1 + 1) \times (1 + 1)$. But it does preserve filtered colimits.
Jun
28
comment Polynomial Functor is ω-Continuous
Sure, $X \mapsto X \times Y$ preserves colimits. That means nothing for $X \mapsto X \times X$.
Jun
28
comment Polynomial Functor is ω-Continuous
Who says $X \mapsto X \times X$ preserves colimits? It doesn't. The claim is not true in absolute generality – it is true for $\mathbf{Set}$ and more generally in $\sigma$-pretoposes, but there is work to be done!
Jun
28
comment Commutativity of direct and inverse limits
See this question.
Jun
27
comment on the coordinate ring of $\mathbb{A}^n \times \mathbb{P}^{m}$
Isn't $\mathbb{A}^n \times \mathbb{P}^m$ basically projective $m$-space over the base $\mathbb{A}^n$? I guess you could use global Proj to make sense of it.
Jun
27
comment coequalizers+pullbacks implies equalizers?
This claim appears in [Paré, 1990, Simply connected limits], where a much more general result is proved concerning the construction of limits for simply connected finite diagrams from pullbacks.
Jun
26
comment Are these parallel theorems from Set Theory and Linear Algebra connected through Category Theory?
The free vector space functor is not dense. It is true that every injective linear map is isomorphic to one in the image of the free vector space functor: use e.g. the Steinitz exchange lemma. The same is true for surjective linear maps: pick a basis for the kernel, extend it to a basis for the whole domain, etc.
Jun
26
comment Initial Algebras in a Category with ω-limits.
Proceed by induction. Of course, $f \circ \delta_0 = \xi_0$ because $0$ is initial. The induction step uses the fact that $F$ preserves $\omega$-colimits.
Jun
26
comment Initial Algebras in a Category with ω-limits.
OK, so you have a homomorphism of $F$-algebras $f : A \to X$. Compose it with the various colimit cocone components $F^n 0 \to A$. What can you say?
Jun
26
comment Initial Algebras in a Category with ω-limits.
You start with an arbitrary diagram, which can't be right. Perhaps you should start with the initial object (of $\mathcal{C}$).
Jun
26
comment Can zero divisors be in the denominator when we localize rings?
There's nothing wrong with localising with respect to $0$ – then you get the trivial ring.
Jun
25
comment Smoothness and field of fractions
You seem to be asking whether generic smoothness implies smoothness?
Jun
25
comment How are the cardinalities of the object images of adjoint functors related?
Well, if the categories in question are concrete, we can just use the usual set-theoretic notion of cardinality. But then we can't say very much about what the functors do, even if we know they are adjoint.
Jun
25
comment If $\mathbb{Z}$ satisfies an identity $\eta$, then every **commutative** ring satisfies $\eta$? And related questions.
As for question 0, it suffices to verify the following: if a (multivariable) polynomial vanishes at all integer points, then it is the zero polynomial. This can be done by induction.
Jun
25
comment If $\mathbb{Z}$ satisfies an identity $\eta$, then every **commutative** ring satisfies $\eta$? And related questions.
Well, if $Z$ is empty (i.e. $T$ has no constants) then every equation is trivially satisfied, including the inconsistent equation $x = y$.
Jun
25
comment Naturality in Yoneda's Lemma
$[\mathcal{C}, \mathbf{Set}]$ is locally small if $\mathcal{C}$ is small; otherwise $[\mathcal{C}, \mathbf{Set}$] may not be locally small (exercise). But if you really want to deal with that, use Mac Lane's $\mathbf{Ens}$ device.