Proving homotopy equivalence of a torus with points removed

Suppose I'd like to show that a Torus with $n$ points removed is homotopy equivalent to a wedge sum of $n+1$ circles. I depict it in a usual way - as a rectangle with $n$ points removed inside. Now it is possible to find a line such that all lines going through those points and parallel to it (the line) are separate. So now I can divide the rectangle in $n+1$ convex strips, each of them containing exactly one rempved point. I know that it is possible to move those removed points homeomorphically in a way that they lie on one line. But how can it be shown in a nice and elegant way? Once it is accomplished it is easy to construct a proper deformation reftraction. Thank you for any hints.

Solution 1:

Instead of choosing lines going through the removed points, consider $n-1$ lines which separate the points into $n$ chambers. Each such punctured chamber deformation retracts onto its boundary circle. (Push along the dotted arrows in the following picture.)

Then you are left with the 1-skeleton of this complex, which is precisely a wedge of $n+1$ circles. (The boundary of the big square, under the quotient map to the torus, is a wedge of two circles. There are $n-1$ circles inside the big square, which are the images of the lines delimiting the chambers under the quotient map, for a total of $2+(n-1) = n+1$ circles.)