The Stacks project

Lemma 90.14.1. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda $ which has (S1).

  1. For $y \to x$ in $\mathcal{F}$ a minimal object in $\mathcal{S}_ y$ maps to a minimal object of $\mathcal{S}_ x$.

  2. For $y \to x$ in $\mathcal{F}$ lying over a surjection $f : B \to A$ in $\mathcal{C}_\Lambda $ every minimal object of $\mathcal{S}_ x$ is the image of a minimal object of $\mathcal{S}_ y$.

Proof. Proof of (1). Say $y \to x$ lies over $f : B \to A$. Let $y' \to y$ lying over $B' \subset B$ be a minimal object of $\mathcal{S}_ y$. Let

\[ \vcenter { \xymatrix{ y' \ar[d] \ar[r] & x' \ar[d] \\ y \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B' \ar[d] \ar[r] & f(B') \ar[d] \\ B \ar[r] & A } } \]

be as in the construction of $f_*$ above. Suppose that $(x'' \to x) \to (x' \to x)$ is a morphism of $\mathcal{S}_ x$ with $x'' \to x'$ lying over $A'' \subset f(B')$. By (S1) there exists $y'' \to y'$ lying over $B' \times _{f(B')} A'' \to B'$. Since $y' \to y$ is minimal we conclude that $B' \times _{f(B')} A'' \to B'$ is an isomorphism, which implies that $A'' = f(B')$, i.e., $x' \to x$ is minimal.

Proof of (2). Suppose $f : B \to A$ is surjective and $y \to x$ lies over $f$. Let $x' \to x$ be a minimal object of $\mathcal{S}_ x$ lying over $A' \subset A$. By (S1) there exists $y' \to y$ lying over $B' = f^{-1}(A') = B \times _ A A' \to B$ whose image in $\mathcal{S}_ x$ is $x' \to x$. So $f_*(y' \to y) = x' \to x$. Choose a morphism $(y'' \to y) \to (y' \to y)$ in $\mathcal{S}_ y$ with $y'' \to y$ a minimal object (this is possible by the remark on lengths above the lemma). Then $f_*(y'' \to y)$ is an object of $\mathcal{S}_ x$ which maps to $x' \to x$ (by functoriality of $f_*$) hence is isomorphic to $x' \to x$ by minimality of $x' \to x$. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06T1. Beware of the difference between the letter 'O' and the digit '0'.