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I'm trying to understand the proof of Theorem 4.23 (Case 1) in Allen Hatcher's Algebraic Topology.

We have a map $f$, for which $f^{-1}(\Delta ^{n+1})$ is a finite union of convex polyhedra, on each of which $f$ is the restriction of a linear surjective map from $\mathbb{R}^{i}$ to $\mathbb{R^{n+1}}$.

And the implication I don't get is this: "For a $q \in \Delta^{n+1}$, $f^{-1}(q)$ is a finite union of convex polyhedra of dimension $\leq i - n -1 $, since $f^{-1} : (\Delta ^{n+1})$ is a finite union of convex polyhedra on each of which $f$ is the restriction of a linear surjection $\mathbb{R}^{i} \rightarrow \mathbb{R}^{n+1}$."

blancket
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Kristoffer
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1 Answers1

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This is as follows. Let $f^{-1}(\Delta^{n+1})=\bigcup_{j=1}^k D_j$, where $D_j$ is a convex polyhedron in $\mathbb{R}^i$. And let $f|_{D_j}=f_j|_{D_j}$, where $f_j:\mathbb{R}^{i}\to \mathbb{R^{n+1}}$ is a linear surjection. Then $f^{-1}(q)=\bigcup_{j=1}^k (D_j\cap f_j^{-1}(q))$. Note that $f_j^{-1}(q)$ is a subspace of dimension $i-n-1$, and hence the intersection $(D_j\cap f_j^{-1}(q))$ is a convex polyhedron of dimension $\le i-n-1$. QED.

Vladimir
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  • Thank you very much for the explanation! Just one last question, which may be trivial, but I don't see it. How to you know that $f_j ^{-1}(q)$ is a subspace of dimension $i-n-1$? – Kristoffer Jun 14 '14 at 10:11
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    $f_j$ is surjective so rank$f_j=n+1$ and hence dim ker$f_j=i-$rank$f_j=i-n-1$. – Vladimir Jun 14 '14 at 10:13
  • Right :) Thanks a lot for taking the time Vladimir. I would upvote your answer if I could. (Too new on this forum). – Kristoffer Jun 14 '14 at 13:19
  • @Kristoffer if you're happy with this answer, feel free to accept it as a satisfactory answer. (the tick underneath the up and down arrows). Welcome to math.se! – Dan Rust Jun 15 '14 at 17:38