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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 …

As far as I can tell you are correct. In fact, any map in the category of simplices coming out of a nondegenerate simplex must be injective. If $a:\Delta^n\to X$ is a nondegenerate simplex and $f:\D …

A monotone map $f:[n]\to[m]$ is partition preserving if for all $i\in[n]$, $i\in t_+$ implies $f(i)\in t_+$ and $i\in t_-$ implies $f(i)\in t_-$. More simply, this means that the inverse image of eve …

Let $X$ be any finite complex such that any map $X\to X$ homotopic to the identity is surjective and for which there is a surjective but nullhomotopic PL map $f:X\to X$ (for instance, $X$ could be $S^ …

This is just a contractible Kan complex. It's equivalent to the same condition where you replace the pairs $(\Delta^k,\partial\Delta^k)$ with all pairs $(A,B)$ where $B\subset A$, since $A$ can be bu …

No, it is not; here is a slightly simpler argument than the one indicated in my comment. As in the answer to the previous question, let $X=\Delta_3/\partial\Delta_3$. It suffices to show that the ca …

You ought to be able to trivialize the bundle over each hemisphere and loop at the transition function on the equatorial S^1 (which is presumably the identity map S^1 \to S^1 acting as rotations on th …

In fact, this holds for the derived category of any connective $E_\infty$ ring spectrum $A$ such that $\pi_i(A)=0$ for $i$ sufficiently large: If $M$ is an $A$-module and $M\otimes_A H\pi_0(A)$ is co …