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The answer is no, because the nerve functor turns an adjoint pair of functors between categories into inverse homotopy equivalences between spaces (this is because of the existence of the unit and cou …

Topologically, you could say that this is true because $K(A,1)$ exists for nonabelian groups $A$. When the action of $G$ on $A$ is trivial, at least, $H^1(G,A)$ should be homotopy classes of maps fro …

For any $n$, there is an $n$-ary smooth function that is not a composition of smooth functions of lower arity; according to this answer to a very similar question, this is due to Vitushkin (at least f …

I would disagree that the hypotheses of the adjoint functor theorem are much stronger than exactness. Left exactness is equivalent to preserving all finite limits, and the hypotheses of the adjoint f …

No such category exists. My original argument for this assumed local smallness and is below the break; here is a simpler argument that does not require local smallness (though it does basically use m …

Here's a counterexample with additive functors on abelian categories. If $A$ is an abelian group, let $F(A)$ denote the subgroup of elements that are divisible by $2$. It is easy to see that $F:Ab\t …

This is true only if $X$ has at most one point. Suppose $i:\mathbf{Top}/X\to \mathbf{Top}$ is a full embedding. Write $Id$ for the terminal object of $\mathbf{Top}/X$, the identity map $X\to X$. Th …

I believe the following is a counterexample. Let $\mathcal{A}$ and $\mathcal{B}$ be closed symmetric monoidal abelian categories such that the unit object $1\in\mathcal{B}$ is projective and let $F:\ …

No such right adjoint exists, even restricted to sober spaces. For simplicity let us take $S=\operatorname{Spec} k$ for some field $k$, and consider the space $X$ having two points, one of which is c …

It's called a groupoid. Given an object $A$, call the degenerate edge from $A$ to itself the identity map at $A$. Given an edge $f:A\to B$, let $f^{-1}:B\to A$ denote the unique edge that fills in a …

This is very far from being true. In fact, a compact space has your property iff it is finite. To prove this, suppose $X$ is infinite and has your property. For each ultrafilter $U$ on $X$ convergi …

Here's a counterexample. Let $X=Y=\mathbb{R}$, and let $Y'$ be a point. Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth map such that $f^{-1}(\{0\})=\{1,1/2,1/3,\dots\}\cup\{0\}$, and let $f'$ map $Y'$ …

Fernando's answer is excellent, but I can't resist mentioning what is perhaps the simplest counterexample to a generalization to your question. As Fernando says, there are counterexamples if you gene …

Here is an expansion of my comment into an answer which I think is very compelling as the "correct" definition for compact Hausdorff spaces, though I agree with others who have said that for general s …

Spheres are (homotopy) cogroups for the same reason that homotopy groups are groups. The comultiplication $S^n \to S^n \vee S^n$ is the map that collapses the equator, the same map that is used to de …