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No, there is not. The cohomology ring $H^*(\mathbb{C}P^{N})$ is a truncated polynomial ring $\mathbb{Z}[x]/(x^{N+1})$, where $x$ has degree $2$. Given any map $f:\mathbb{C}P^{N}\to\mathbb{C}P^{N}$, …

I think this is the closest thing there is to what you are looking for. Let $G$ be any injective (i.e., divisible) abelian group. Then there is a spectrum $I_G$ representing the cohomology theory $X …

Here's a partial positive answer, in response to your comment. Suppose $X=\Sigma X'$ and $Y=\Sigma Y'$ are suspensions of connected spaces whose homology is finitely generated in each degree. Then i …

This can fail badly even if the closure of each cell only intersects lower-dimensional cells. For instance, take an open disk together with some subset of the boundary that has infinitely many connec …

As I said in my comment, Ghrist's theorem is stated in too much generality to possibly be true. But if, say, you restrict to requiring $K$ to be a finite CW complex, here's a counterexample to your q …

Yes, by Yoneda's lemma. To generalize a bit, you have a group object $G$ in some category with finite products, and you ask whether the functor it represents is pointwise abelian. This is the same a …

The sphere spectrum, representing stable cohomotopy, is irreducible (you can see this, for instance, from the fact that its homology is irreducible and any summand would be connective and thus would h …

It is true. Let's first assume that $X$ is a finite complex. Then by Alexander duality, $H^2(X;\mathbb{Z})=H_0(S^3\setminus X,\mathbb{Z})$ must be torsion-free. Now the squaring map $H^1(X;\mathbb{ …

This is closely related to a question of mine, which was motivated by wondering whether the mapping cylinder of a homotopy equivalence is a fibration over an interval. The counterexample given there …

Any topological group $G$ has a classifying space, whose loopspace is a (homotopy) group which is homotopy equivalent to $G$ in a way that preserves the group structure. More generally, if $G$ is an …

Yes, we can conclude that either $q$ is an equivalence or $X$ is contractible. Since any cycle lives in a compact subset of $X$, $q$ will also induce surjections on homology. It follows that $X$ is …

Here is a partial affirmative answer using mod 2 Steenrod operations; the simplest case of this (for $n$ and $k$ even) is just a correction of the slightly incorrect answer originally posted by Włodzi …

If $\pi_2(M\setminus\{p\})=0$ then $M$ is simply connected (and hence $S^3$ by the Poincare conjecture). To see this, consider the universal cover $q:U\to M$ and let $V=q^{-1}(M\setminus \{p\})$. Th …

The mod 2 cohomology $H^*(K(\mathbb{Z}/2,n))$ is freely generated over the Steenrod algebra by the canonical class $\iota\in H^n$ in degrees $*\leq 2n$, and the only relation introduced in degree $2n+ …

This might be a rather different direction than you're interested in, but analogues of Quillen's theorem have been studied for graded cocommutative Hopf algebras. These generalizations were originall …