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Let $I$ be the unit interval, and define the torus using the usual identifications on $I \times I$. I've shown that $I \times I - \{x\}$ (where $x$ is a point not on the boundary of $I \times I$) is homotopy equivalent to its boundary (and that it even retracts onto its boundary). Let $p: I \times I \rightarrow I \times I /\sim$ be the canonical quotient map. Using the retraction above, call it $r$, is there a way I can explicitly show that $p(I \times I - \{x\})$ is homotopy equivalent to $p(\text{boundary of } I \times I)$, i.e., that torus minus a point to a figure-eight?

More generally, if $A \subset X$, and $X$ is homotopy equivalent to $A$ (or $X$ retracts to $A$), for which maps $p$ can I say that $p[X]$ is homotopy equivalent to $p[A]$?

Hints appreciated!

user121410
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  • Have you tried checking if the homotopy equivalence induced by the retraction descends nicely to the quotient? – Elchanan Solomon Feb 02 '14 at 20:48
  • @IsaacSolomon Sorry, newbie question, but what does "descending nicely to the quotient" mean? I think I know, but I'd like to make sure first! – user121410 Feb 02 '14 at 20:56
  • Pretty much exactly what Daniel Rust has explained in his answer, i.e. that there is a continuous map $\tilde{r}$ that is sort of the "quotient map" of the original map. – Elchanan Solomon Feb 02 '14 at 21:23

1 Answers1

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Let $r\colon I\times I\setminus\{p\}\to \partial (I\times I)$ be the retraction you mentioned. Can you show that there exists a retraction $\tilde{r}\colon (I\times I\setminus\{p\})/{\sim}\to \partial (I\times I)/{\sim}$ which make the following square commute?

$$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} I\times I\setminus\{p\} & \ra{r} & \partial (I\times I) \\ \da{q} & & \da{q|_{\partial(I\times I)}} \\ (I\times I\setminus\{p\})/{\sim} & \ras{\tilde{r}} & \partial (I\times I)/{\sim} \\ \end{array} $$

Dan Rust
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