In my lecture notes, it says that the mapping torus of a map $\sigma: S^1 \longrightarrow S^1$, defined by $T(\sigma) := S^1 \times I /_{(x,\;1) \sim (\sigma(x), \;0 )}$, for which I know a pushout diagram $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^1 \times \{0,\;1\} & \ra{\hspace{0.35cm} f \hspace{0.35cm}} & S^1 \\ \da{inc} & & \da{g}\\ S^1 \times I & \ras{can} & T(\sigma)\\ \end{array} $$ where $$ \begin{align} f(x,\;0) = x,\quad f(x, \;1) = \sigma(x) \\ g(x) = [(x,\; 0)] \\ \end{align} $$ and $can$ denotes the canonical projection. But it supposedly has another pushout diagram $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^1 & \ra{\hspace{0.35cm}\varphi\hspace{0.35cm}} & S^1 \vee S^1 \\ \da{inc} & & \da{}\\ D^2 & \ras{\hspace{0.8cm}} & T(\sigma)\\ \end{array} $$ . I can't wrap my head around this right now; meaning I fail to find a meaningful image in my head of how to glue the disk to the wedge in order to obtain something like a torus. I see the wedge as in here:

The red circle and the pink circle intersect at the identification point of the wedge

but how do I glue $D^2$ on this to obtain a mapping torus for $\sigma$? Can someone (help me) make $\phi$ explicit? I'm studying for an exam, this is not homework. The lecture notes simply don't say anything about how the pushout is constructed. Please be explicit and avoid 'words about images', if possible, I'd like to have formulas.
$D^2 \cong I^2$ might help here, I guess...

P.s.: This is not central to the topic right now, either. What I want to prove is just part of another proof concerning the homological degree...


1 Answer 1


Let $\iota_0, \iota_1:\;S^1 \hookrightarrow S^1 \vee S^1$ denote the canonical inclusions. Define $\varphi: S^1 \longrightarrow S^1 \vee S^1$ by $$ \begin{align} \varphi(e^{2\pi it}):= \begin{cases} \iota_0\sigma(e^{2\pi i (4t)})\;, & t\in[\,0,\,1/4\,] \\ \iota_1 e^{2\pi i (4t - 1)}\;, & t\in[\,1/4,\,1/2\,] \\ \iota_0 e^{2\pi i (4(1-t) - 2)}\;, & t\in[\,1/2,\,3/4\,] \\ \iota_1\sigma(e^{2\pi i (4(1-t) - 3)})\;, & t\in[\,3/4,\,1\,] \\ \end{cases} \end{align} $$ One checks that this is indeed well-defined. The following pushout will then yield a space homemomorphic to the mapping torus: $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^1 & \ra{\hspace{0.35cm} \varphi \hspace{0.35cm}} & S^1 \vee S^1 \\ \da{inc} & & \da{}\\ D^2 & \ras{\hspace 0.9cm} & D^2 \bigcup_\varphi (S^1 \vee S^1)\\ \end{array} $$

Also note: If the corresponding pushout is produced using the representations $\partial I^2$ and $I^2$ instead of $S^1 = \partial D^2$ and $D^2$ under the obvious homeomorphy, a pushout could look like this: $$ \begin{align} \varphi'(t_1,t_2):= \begin{cases} \iota'_0\sigma(e^{2\pi i (t_2)})\;, & (t_1, t_2)\in \{1\} \times [\,0,\,1\,] \\ \iota'_1 e^{2\pi i (t_1)}\;, & (t_1, t_2)\in [\,0,\,1\,] \times \{1\} \\ \iota'_0 e^{2\pi i (t_2)}\;, & (t_1, t_2)\in \{0\} \times [\,0,\,1\,] \\ \iota'_1\sigma(e^{2\pi i (t)})\;, & (t_1, t_2)\in[\,0,\,1\,] \times \{0\} \\ \end{cases} \end{align} $$ $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} \partial I^2 & \ra{\hspace{0.35cm} \varphi \hspace{0.35cm}} & S^1 \vee S^1 \\ \da{inc} & & \da{}\\ I^2 & \ras{\hspace 0.9cm} & I^2 \bigcup_\varphi (S^1 \vee S^1)\\ \end{array} $$ where $\iota'_0, \iota'_1:\;\partial I^2 \hookrightarrow S^1 \vee S^1$.
This version of a pushout might be more in line with what most professors would 'draw' and I find it easier to visualize the gluing with such a square, compared to gluing a round disk. On the other hand, the first pushout makes it much clearer, how the circle is mapped into the wedge.

The pushout diagrams would commute through usage of the homeomorphy $$ \begin{align} \phi:&\;D^2 \longrightarrow S^1 \\ & x \mapsto \begin{cases} 0,\;x=0 \\ \frac{\|x\|_2 x}{\|x\|_1},\; x \ne 0 \end{cases} \end{align} $$ which restricts to a homeomorphism between the boundaries.


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