Deformation Retract of Open Square With 3 Points Removed

Question

Consider the open unit square with three points removed (say $a, b, c$) $S=(0,1)\times(0,1)-\left\{a, b, c\right\}$. Is a bouquet of three circles (wedge sum of three circles) a deformation retract of $S$? If so, why?

Answer 1

Pick any of the points remaining in the square. Draw loops starting at that point, encircling a single one of the removed points, without self intersections, and without intersecting the other two loops.

Since the interior of the loops have a removed point each, then they can be retracted to the points of the loop. The exterior of the concatenation of all the loops can be retracted to the points of the loops as well.

Answer 2

Take the square with one point removed. You can certainly radiallyretract to the boundary to just get $S^1$.

For $2$ points removed, put one at $(1/3,1/2)$ and $(2/3,1/2)$ (which I'll include just for concreteness.

Using the vertical line at $x=1/2$, you can radially retract each "half of the square" onto the boundary of the "half square."

You can do this for $n$ points removed, by spacing them evenly and considering "$1/3$ squares" and so forth.