The Stacks Project


Tag 017R

Chapter 14: Simplicial Methods > Section 14.18: Splitting simplicial objects

Lemma 14.18.2. Let $U$ be a simplicial set. Then $U$ has a unique splitting with $N(U_m)$ equal to the set of nondegenerate $m$-simplices.

Proof. From the definition it follows immediately, that if there is a splitting then $N(U_m)$ has the be the set of nondegenerate simplices. Let $x \in U_n$. Suppose that there are surjections $\varphi : [n] \to [k]$ and $\psi : [n] \to [l]$ and nondegenerate simplices $y \in U_k$, $z \in U_l$ such that $x = U(\varphi)(y)$ and $x = U(\psi)(z)$. Choose a right inverse $\xi : [l] \to [n]$ of $\psi$, i.e., $\psi \circ \xi = \text{id}_{[l]}$. Then $z = U(\xi)(x)$. Hence $z = U(\xi)(x) = U(\varphi \circ \xi)(y)$. Since $z$ is nondegenerate we conclude that $\varphi \circ \xi : [l] \to [k]$ is surjective, and hence $l \geq k$. Similarly $k \geq l$. Hence we see that $\varphi \circ \xi : [l] \to [k]$ has to be the identity map for any choice of right inverse $\xi$ of $\psi$. This easily implies that $\psi = \varphi$. $\square$

    The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 1784–1788 (see updates for more information).

    \begin{lemma}
    \label{lemma-splitting-simplicial-sets}
    Let $U$ be a simplicial set. Then $U$ has a unique splitting
    with $N(U_m)$ equal to the set of nondegenerate $m$-simplices.
    \end{lemma}
    
    \begin{proof}
    From the definition it follows immediately, that if there is a
    splitting then $N(U_m)$ has the be the set of nondegenerate simplices.
    Let $x \in U_n$. Suppose that there are surjections $\varphi : [n] \to [k]$
    and $\psi : [n] \to [l]$ and nondegenerate simplices
    $y \in U_k$, $z \in U_l$ such that $x = U(\varphi)(y)$
    and $x = U(\psi)(z)$. Choose a right inverse $\xi : [l] \to [n]$
    of $\psi$, i.e., $\psi \circ \xi = \text{id}_{[l]}$.
    Then $z = U(\xi)(x)$. Hence $z = U(\xi)(x) = U(\varphi \circ \xi)(y)$.
    Since $z$ is nondegenerate we conclude that $\varphi \circ \xi :
    [l] \to [k]$ is surjective, and hence $l \geq k$. Similarly
    $k \geq l$. Hence we see that $\varphi \circ \xi : [l] \to [k]$
    has to be the identity map for any choice of right inverse
    $\xi$ of $\psi$. This easily implies that $\psi = \varphi$.
    \end{proof}

    Comments (0)

    There are no comments yet for this tag.

    There are also 3 comments on Section 14.18: Simplicial Methods.

    Add a comment on tag 017R

    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 lower-right corner).

    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?