I am reviewing some technical results from the fourth chapter of Hatcher's Algebraic Topology. In the proof of PL-approximation (Lemma 4.10), we let $B_1,B_2\subset e^k$ denote the balls of radius 1 and 2 (respectively) centered at the origin under some fixed identification $h\colon e^k\to\mathbb R^k$. We then define suitable polyhedra $K_1,K_2\subset f^{-1}(B_2)$ which are used throughout the proof. However, I see no reason to exclude the possibility that $f^{-1}(B_1) = f^{-1}(B_2)$. Indeed, $f$ is not assumed to be surjective, so why not? We could have $\mathrm{im} f\cap B_1 = \mathrm{im} f\cap B_2$. This detail is important for two reasons:
- To ensure that $K_1$, which is defined as a union of cubes of diameter $<\varepsilon$ (for some suitable $\varepsilon > 0$) intersecting $f^{-1}(B_1)$, is contained in $f^{-1}(B_2)$.
- More concretely, Hatcher says (at the bottom of page 350, just above the picture) that we may assume that $\varepsilon$ is less than half of the distance between $f^{-1}(B_1)$ and $I^n\setminus f^{-1}(\mathrm{int} B_2)$. However, when $f^{-1}(B_1) = f^{-1}(B_2)$, this distance is zero, and we obviously need $\varepsilon > 0$.
The proof can be adjusted by picking balls of radii depending on $f$, and then there should be no problem. Am I right to think that some adjustment like this is necessary?
My book is the 2015 reprinting. This updated online version has the same problem (if it is a problem, which I suspect).
EDIT. I claim above that the radii can be adjusted to fix the proof. But, even with adjusted radii $r$ and $s$, there are still two problems with the approach in Hatcher:
- What if $f^{-1}(B_r) = f^{-1}(B_s) = \varnothing$? (This is easily fixed by considering balls whose center are in the image of $f$.)
- What if $f$ is constant? Then $f^{-1}(B_r) = f^{-1}(B_s)$ for balls of arbitrary radius and center. This suggests that, at the very least, the proof should treat constant $f$ separately. (Of course constant $f$ are already piecewise linear, so here the proposition holds trivially.)
Since we have $B_1 \subset \text{int } B_2 \subset B_2$, it follows that we have the same sequence of inclusions after applying $f^{-1}$. If $f^{-1}(B_1) = f^{-1}(B_2)$, then both are also equal to $f^{-1}(\text{int } B_2)$, let $A\subset I^n$ be this common subset. Then, $A$ is closed, and also open, so it follows by connectivity that $A = I^n$ or $A=\emptyset$. If $f$ has image outside of the cell, there is nothing to prove, so we can assume that $A\neq \emptyset$ by choosing identifications as you mention, and if $A = I^n$, then a convex homotopy to a constant map (which is PL), satisfies the lemma. Thus, it is safe to continue assuming that the inclusions $\emptyset \subset f^{-1}(B_1) \subset f^{-1}(\text{int } B_2) \subset I^n$ are proper.