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David Roberts
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Take $\mathbb{R}^2$ and remove the points $(n,0)$ for $n\in \mathbb{Z}$ along the $x$-axis. This is homotopically equivalent to the wedge of countably many circles. Take the union of a (clockwise) loop around every deleted point, based at (0,-1), and chosen to be disjoint from each other. This is the image of the countable wedge and the described space deformation retracts onto it, making them homotopically equivalent.

Alternatively, delete only the points at positive integers along the $x$-axis. Then take loops based at $(-1,0)$ which are circles of increasing radius, including one more deleted point for each new circle (this is a sort of reflected Hawaiian earring). The inclusion of the subspace which is the union of all the circles is a homotopy equivalence between the countable wedge and this new space, and metrises the countable wedge nicely.

Take $\mathbb{R}^2$ and remove the points $(n,0)$ for $n\in \mathbb{Z}$ along the $x$-axis. This is homotopically equivalent to the wedge of countably many circles. Take the union of a (clockwise) loop around every deleted point, based at (0,-1), and chosen to be disjoint from each other. This is the image of the countable wedge and the described space deformation retracts onto it, making them homotopically equivalent.

Alternatively, delete only the points at positive integers along the $x$-axis. Then take loops based at $(-1,0)$ which are circles of increasing radius, including one more deleted point for each new circle (this is a sort of reflected Hawaiian earring). The inclusion of the subspace which is the union of all the circles is a homotopy equivalence between the countable wedge and this new space, and metrises the countable wedge nicely.

Take $\mathbb{R}^2$ and remove the points $(n,0)$ for $n\in \mathbb{Z}$ along the $x$-axis. This is homotopically equivalent to the wedge of countably many circles. Take the union of a (clockwise) loop around every deleted point, based at (0,-1), and chosen to be disjoint from each other. This is the image of the countable wedge and the described space deformation retracts onto it, making them homotopically equivalent.

Alternatively, delete only the points at positive integers along the $x$-axis. Then take loops based at $(-1,0)$ which are circles of increasing radius, including one more deleted point for each new circle (this is a sort of reflected Hawaiian earring). The inclusion of the subspace which is the union of all the circles is a homotopy equivalence between the countable wedge and this new space.

Source Link
David Roberts
  • 35.4k
  • 11
  • 124
  • 349

Take $\mathbb{R}^2$ and remove the points $(n,0)$ for $n\in \mathbb{Z}$ along the $x$-axis. This is homotopically equivalent to the wedge of countably many circles. Take the union of a (clockwise) loop around every deleted point, based at (0,-1), and chosen to be disjoint from each other. This is the image of the countable wedge and the described space deformation retracts onto it, making them homotopically equivalent.

Alternatively, delete only the points at positive integers along the $x$-axis. Then take loops based at $(-1,0)$ which are circles of increasing radius, including one more deleted point for each new circle (this is a sort of reflected Hawaiian earring). The inclusion of the subspace which is the union of all the circles is a homotopy equivalence between the countable wedge and this new space, and metrises the countable wedge nicely.