Lemma 14.31.7. Let $f : X \to Y$ be a homomorphism of simplicial abelian groups which is termwise surjective. Then $f$ is a Kan fibration of simplicial sets.
Proof. Consider a commutative solid diagram
\[ \xymatrix{ \Lambda _ k[n] \ar[r]_ a \ar[d] & X \ar[d] \\ \Delta [n] \ar[r]^ b \ar@{-->}[ru] & Y } \]
as in Definition 14.31.1. The map $a$ corresponds to $x_0, \ldots , \hat x_ k, \ldots , x_ n \in X_{n - 1}$ satisfying $d_ i x_ j = d_{j - 1} x_ i$ for $i < j$, $i, j \not= k$. The map $b$ corresponds to an element $y \in Y_ n$ such that $d_ iy = f(x_ i)$ for $i \not= k$. Our task is to produce an $x \in X_ n$ such that $d_ ix = x_ i$ for $i \not= k$ and $f(x) = y$.
Since $f$ is termwise surjective we can find $x \in X_ n$ with $f(x) = y$. Replace $y$ by $0 = y - f(x)$ and $x_ i$ by $x_ i - d_ ix$ for $i \not= k$. Then we see that we may assume $y = 0$. In particular $f(x_ i) = 0$. In other words, we can replace $X$ by $\mathop{\mathrm{Ker}}(f) \subset X$ and $Y$ by $0$. In this case the statement become Lemma 14.31.6. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: