Proposition

Construct an explicit deformation retraction of the torus with one point deleted onto a graph consisting of two circles intersecting in a point, namely, longitude and meridian circles of the torus.

Solution

First, we will define a torus. Let $X = \{ (x, y) \in \mathbb{R}^2 \mid \abs{x} \leq 1 \land \abs{y} \leq 1 \}$. Define $f_1, f_2$ such that

Let $p = f_1 \circ f_2$. Let $Y = p(X) = \{ (x, y) \in \mathbb{R}^2 \mid \abs{x} \leq 1, \abs{y} \leq 1, x \ne -1, y \ne -1 \}$. Then $Y$ with the quotient topology induced by $p$ is a torus.

Next, we will define a function on $X$ because it is easier to see continuity when dealing with real functions, rather than functions on quotient spaces. Let $g((x, y), t): X \times I \rightarrow X$ be defined such that

where $\phi(t) = (1 - t) + t \max \{ \abs{x}, \abs{y} \}$.

Then $g$ is continuous since each component function is a continuous real function.

Lastly, we will define a deformation retraction. The composition $p \circ g$ maps $X \times I$ into $Y$ continuously because both $p$ and $g$ are continuous.

Let $(x, y) \in Y, t \in I$ be given. We claim that $(x, y) \maps g((x, y), t)$ is constant on $p^{-1}(\{ (x, y) \})$.

• If $x \ne -1$ and $y \ne -1$, then $p^{-1}(\{ (x, y) \}) = \{ (x, y) \}$.
• If $x = -1$ and $y \ne -1$, then $p^{-1}(\{ (x, y) \}) = \{ (1, y), (-1, y) \}$.
• If $x \ne -1$ and $y = -1$, then $p^{-1}(\{ (x, y) \}) = \{ (x, 1), (x, -1) \}$.
• If $x = -1$ and $y = -1$, then $p^{-1}(\{ (x, y) \}) = \{ (1, 1), (-1, 1), (1, -1), (-1, -1) \}$.

TODO