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Why is the torus with a point missing homotopy equivalent to figure-8-space?

$T^2 \setminus \{p\} \simeq S^1 \vee S^1$

iwriteonbananas
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    Consider the fundamental square $[0, 1]^2$ of the torus, with opposite edges identified preserving order. Chuck a point out of the interior of the square. Note that everything deformation retracts onto the boundary. Now do the identifications to verify what you get is really $S^1 \vee S^1$. – Balarka Sen Feb 21 '15 at 10:59
  • http://math.stackexchange.com/questions/661219/homotopy-equivalence-from-torus-minus-a-point-to-a-figure-eight – Dario Feb 21 '15 at 11:22

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