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16h

revised 
Convergence of a sum with the ranks of homotopy groups
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17h

comment 
Convergence of a sum with the ranks of homotopy groups
If $F$ is simply connected and a finite CWcomplex, then $I(F)<\infty$ iff $F$ has only finitely many nontrivial rational homotopy groups (see Theorem 2.33 here, for instance; I'm not sure whether the simply connected hypothesis is really necessary). If you don't demand $F$ to be a finite CWcomplex, the question seems hopeless in full generality (for instance, $F$ could be an arbitrary product of spheres of different dimensions). Do you have any particular motivation for looking at $I(F)$? 
17h

revised 
Convergence of a sum with the ranks of homotopy groups
edited tags 
May 22 
answered  $\mathcal S'(\mathbb R^d)$ is separable 
May 22 
comment 
$\mathcal S'(\mathbb R^d)$ is separable
Why is $S'(\mathbb{R}^d)$ isomorphic to $s'$? 
May 17 
comment 
Vectorisation of a category
If you allow objects of $\operatorname{Vec}(C)$ to be multisets rather than just sets of objects of $C$, then you can describe this as just freely adjoining coproducts to the $\mathbb{R}$linearization of your category. 
May 14 
comment 
Is there a (satisfying) proof that cellular cohomology is isomorphic to simplicial cohomology that doesn't use relative cohomlogy?
By the way, the proof in Hatcher is exactly the same as the spectral sequence proof Ben alludes to, just without using the word "spectral sequence". 
May 11 
comment 
Partition refinement of a clopen covering in $\Box (\omega+1)^\omega$
Can you not just take the $U_f$'s for $f$ such that if $f(n)<\omega$, then $f(n)<n$? 
May 10 
comment 
When does a function space allow for point evaluations?
It seems that before seeking "necessary and sufficient conditions", you should seek a precise definition of what it even means to have point evaluations for "generalized functions" or functions that are probably a priori defined only up to null sets. 
May 9 
comment 
Is there a 'largest' secondorder categorical axiomatization of set theory, extended from ZFC2
Is there any justification for believing that if "$\ZFC+$ there is a model of $T$" is consistent then so is "$\ZFC+$ there is a model of $T$ and a larger inaccessible cardinal", other than the fact that no one has ever found a $T$ for which this seems to be false? 
May 7 
comment 
Is there a Leibnizian model with no definable elements, in a finite language?
For an alternate argument for nondefinability (for the collection of all noneventually periodic sequences), you can take an ultrapower and construct automorphisms. Explicitly, if $x$ is a sequence and $x_n$ is a sequence of distinct sequences that converge to $x$ (pointwise), then there is an automorphism of the ultrapower that exchanges $x$ and $[x_n]$ (just exchange the entire orbits of $x$ and $[x_n]$ with respect to shifts (both left and right); these orbits are in bijection as long as $x$ and the $x_n$ are not eventually periodic). 
May 7 
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Is there a Leibnizian model with no definable elements, in a finite language?
If you throw out all sequences that are eventually periodic I think you can avoid problems with equality. 
May 7 
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Graph of graph homomorphisms
@ViditNanda: The Cartesian product of graphs, which is adjoint to the Hom described in the question. 
May 6 
comment 
Is every abelian group a colimit of copies of Z?
@TimCampion: That group is not torsionfree because $2(xy)=0$. 
May 6 
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Is every abelian group a colimit of copies of Z?
@QiaochuYuan: Perhaps an easier way to think about what happens in the torsionfree case once you've reduced to a single equivalence class is to tensor everything with $\mathbb{Q}$. Doing this turns your diagram into a commuting diagram where every object is $\mathbb{Q}$ and every map is an isomorphism, so the colimit is clearly $\mathbb{Q}$. Since your group was torsionfree, it injects into its tensor product with $\mathbb{Q}$ and is hence a subgroup of $\mathbb{Q}$. 
May 5 
comment 
The number of ideals in a ring
If $R=k\oplus V$ with $V^2=0$, where $V$ is a sufficiently largedimensional vector space over a finite field $k$, then $I_R$ will be much larger than $R$. 
May 3 
awarded  Enlightened 
May 3 
awarded  Nice Answer 
May 3 
comment 
Axiomatic approach to means
This question is related and has a bit of discussion of past work on axiomatics in the comments. 
May 1 
comment 
Graph of graph homomorphisms
I suspect that for finite $G$, this holds for most graphs $H_1$ with $V(H_1)\ggV(G)$ and $H_2=G\square H_1$ (heuristically, for random $H_1$ there should be no maps $H_1\to G\square H_1$ other than the obvious ones). For infinite $G$, the question may involve some nontrivial set theoryfor instance, if $G$ is a complete infinite graph, it is easy to see that $H_1$ and $H_2$ must have greater cardinality than $G$. 