# A is Homotopy Equivalent to A^op via Functors

Let $\mathcal{A}$ be a small category and $\mathbf{B}\mathcal{A}$ its geometric realisation. It is evident that $\mathbf{B}\mathcal{A}$ and $\mathbf{B}\mathcal{A}^\circ$ are homotopy equivalent, and in fact homeomorphic. However, can we find functors that realise this equivalence? This post summarises some informal notes I have written on this following D. Quillen's paper Higher Algebraic K-Theory: I", so grab the notes or read the summary below:

Given any functor $f:\mathcal{A}\to\mathcal{B}$, and an object $B\in\mathcal{B}$, we can consider the category $f^{-1}(B)$ consisting of objects $A\in \mathcal{A}$ such that $f(A) = B$. The morphisms of $f^{-1}(B)$ are defined to be all the morphisms that map to $1_B$ under $f$. Let us apply this to the following situation:

Given any small (or skeletally small) category $\mathcal{A}$, we can construct another category $S(\mathcal{A})$ as follows: the objects of $S(\mathcal{A})$ are the arrows $X\to Y$ of $\mathcal{A}$, and a morphism $(X\to Y)\to (X'\to Y')$ is a pair of morphisms $X'\to X$ and $Y\to Y'$ in $\mathcal{A}$ making the obvious square commute. Now, we can consider the functor $s:S(\mathcal{A})\to \mathcal{A}$ sending the object $X\to Y\in S(\mathcal{A})$ to the object $X\in\mathcal{A}$.
More »

# Homotopy on Chain Complexes

Posted by Jason Polak on 29. May 2013 · Write a comment · Categories: homological-algebra · Tags: ,

Let $R$ be any associative ring with unit and $A$ an $R$-module. If $P$ is a projective module and $A\to P\to A = 1_A$, is $A$ necessarily projective?