TITLE: How do you show that $S^{\infty}$ is contractible?
QUESTION [42 upvotes]: Here I mean the version with all but finitely many components zero.

REPLY [8 votes]: A very nice proof is via classifying spaces of categories. It goes as follows: Take the “walking isomorphism“ category $J$, that is the unique category with two isomorphic objects and four morphisms in total. The geometric realization of its simplicial nerve is exactly the infinity-sphere! (The non-degenerate $n$-simplices are given by the functors $[n] \to J$ starting at either $0$ or $1$ and then going back and forth using the isomorphism to produce upper and lower hemisphere of the standard cell-decomposition.) Then you conclude by observing that this category is equivalent to the terminal category and thus, as nerve and realization send natural transformations to simplicial, respectively topological homotopies, $S^\infty$ is contractible.