Let me restate the theorem:

Let $F\colon\mathcal{C}\to\mathcal{D}$ be a functor. If $F\downarrow x$ is contractible for every $x\in\operatorname{Ob}(\mathcal{D})$, then $F$ is a homotopy equivalence.

Now let $\mathcal{C}$ be the category with two objects $x$, $y$, and two non-identitymorphisms $f_1, f_2\in\mathcal{C}(x,y)$, $\mathcal{D}$ be the category with two objects $\tilde x$, $\tilde y$ and one non-identitymorphism $\tilde f\in\mathcal{D}(\tilde x,\tilde y)$, and $F\colon\mathcal{C}\to\mathcal{D}$ be the obvious functor. Then

  • $F\downarrow \tilde x$ is the category with one object $(x,\operatorname{id}_{\tilde x})$ and no non-identitymorphisms. In particular, $\left|N(F\downarrow \tilde x)\right|\cong *$, so $F\downarrow \tilde x$ is contractible.
  • $F\downarrow \tilde y$ is the category with two objects $(x,\tilde f), (y, \operatorname{id}_{\tilde y})\in\operatorname{Ob}(F\downarrow \tilde y)$ and one non-identitymorphism $\tilde f\colon (x,\tilde f)\to (y, \operatorname{id}_{\tilde y})$. In particular, $\left|N(F\downarrow \tilde y)\right|\cong [0,1]$, so $F\downarrow \tilde y$ is contractible.

But $\left|N(\mathcal{C})\right|\cong S^1$ and $\left|N(\mathcal{D})\right|\cong [0,1]$, so $F$ can't be a homotopy equivalence.

I'm quite sure I didn't disproof Quillens Theorem A using one of the easiest examples one could think of, but I just can't find my mistake...

  • 1
    $\begingroup$ There are two morphisms $(x,\tilde{f})\to(y,\mathrm{id}_{\tilde{y}})$ : $f_1$ and $f_2$, so $F\downarrow \tilde{y}$ is isomorphic to $\mathcal{C}$, and not contractible. $\endgroup$ – Olivier Bégassat Nov 6 '13 at 13:59

I think you just confused what morphisms in $F\downarrow x$ are : given two objects $(a,\phi)$ and $(b,\psi)$ (that is two objects $a,b\in\mathcal C$ and two morphisms $\phi:Fa\to x$ and $\psi:Fb\to x$ in $\mathcal{D}$), a morphism between them is a morphism $\gamma:a\to b$ such that $\psi\circ F\gamma=\phi$. In this case, there are two morphisms $(x,\tilde{f})\to(y,\mathrm{id_{\tilde{y}}})$ : $f_1$ and $f_2$. Both $\gamma=f_1$ and $\gamma=f_2$ morphisms satisfy the equation $\mathrm{id_{\tilde{y}}}\circ F\gamma=\tilde f$.


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