ScalingStacks

[05VP]
  1. (1)

    Given f∈map⁡(x,y)f\in\map(x,y) and g∈map⁡(y,z)g\in\map(y,z), and g∘fg\circ f the result of a composition, then (g∘f)∗∼g∗∘f∗(g\circ f)_{*}\sim g_{*}\circ f_{*} and (g∘f)∗∼f∗∘g∗(g\circ f)^{*}\sim f^{*}\circ g^{*}.

  2. (2)

    Given x∈ob⁡Wx\in{\operatorname{ob}}W then (idx)∗∼(idx)∗∼idmap⁡(x,x)(\id_{x})_{*}\sim(\id_{x})^{*}\sim\id_{\map(x,x)}.

[05VQ]

Proof. To prove (1), let k∈map⁡(x,y,z)k\in\map(x,y,z) be a composition of ff and gg which results in a composite g∘fg\circ f. To show that (g∘f)∗∼g∗∘f∗(g\circ f)_{*}\sim g_{*}\circ f_{*}, it suffices to show that both sides of the equation are equal (in the homotopy category of spaces) to the zig-zag

map⁡(w,x)→{k}×1map⁡(x,y,z)×map⁡(w,x)←∼map⁡(w,x,y,z)→map⁡(w,z).\map(w,x)\xrightarrow{\{k\}\times 1}\map(x,y,z)\times\map(w,x)\xleftarrow{\sim}\map(w,x,y,z)\rightarrow\map(w,z).

The proof that (g∘f)∗∼f∗∘g∗(g\circ f)^{*}\sim f^{*}\circ g^{*} is similar.

The proof of (2) is straightforward. ∎

Original mathematics by the credited authors. Source collection and HTML conversion remain in progress.

Charles Rezk

Original source: arXiv:math/9811037v3

Original source page 25

Original source · math/9811037v3