3
$\begingroup$

I have a question from Hatcher's Algebraic topology Chapter 0, problem 6: "Let $X$ be the subspace of $\mathbb{R}^2$ consisting of the horizontal segment $[0,1]\times\{0\}$ together with the vertical segments $\{r\}\times[0,1-r]$ for $r$ a rational number in $[0,1].$ Show that $X$ deformation retracts to any point in the segment $[0,1]\times\{0\}$, but not to any other point."

I have a solution to the first part: First, $X$ deformation retracts onto $[0, 1] \times \{0\}$: Take the family of functions $$f_t(r, x) = (r, (1-t)x).$$ It's easy to see that this is a deformation retraction onto $[0, 1] \times \{0\}$. Likewise, the family $$ g_t(x, 0) = ((1-t)x + ta, 0) $$ is a deformation retraction onto the point $(a, 0)$. Hence, composing these, we have the family $$ h_t = \begin{cases} f_{2t} & \text{ for $0\leq t\leq \frac{1}{2}$} \\ g_{2t-1} & \text{ for $\frac{1}{2}\leq t\leq 1$}, \end{cases} $$ which is a deformation contraction onto $(a, 0)$.

But I don't see how these are the only points. For some $(r, x)$, can't we just deformation retract every vertical line $\{q\} \times [0, 1-q]$, $q\neq r$ down to $[0, 1] \times \{0\}$, then deformation retract $[0, 1] \times \{0\}$ to $(r, 0)$, and then deformation retract the rational line $\{r\} \times [0, 1-r]$ to $(r, x)$? What's wrong with this deformation retraction?

Thanks so much.

$\endgroup$
2
$\begingroup$

This is a rather informal argument for why (it is not a proof which needs more care - it's at best a sketch).

Note that you can't do the process you describe because you can't move the point you're wanting to deformation retract on to, and if you can't move that point, then you can't move the points in a small neighbourhood around it 'too far'. Because the rationals are dense, that means there will be points on other lines that won't be able to push all the way down to the base line without 'breaking' continuity in some way. If the set of $r$ were chosen from some discrete subset of $\mathbb{R}$ it would be fine. It's precisely because $\mathbb{Q}$ is dense in $\mathbb{R}$ that we can't move individual lines without considering those which are close to it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.