Torus with a point deleted is not a retract of the torus. Show that the 2-torus with a deleted point $T\setminus \{ x_0\}$ is not a retract of $T$.  
I know that we can prove the torus with a point removed deformation retracts to the wedge of two circles. 
However, I don't know how to show $T\setminus \{ x_0\}$ is not a retract of $T$.  And I haven't been able to find anything online. 
Please help.
 A: EDIT: Sigh... the torus is compact, but $T\backslash \{x_0\}$ isn't. $\blacksquare$
Nevertheless, I'm gonna keep the answer below.

If $A \subset B$ is a retract of $B$, then the induced map (by inclusion) on the fundamental group is injective. This is seen readily from functoriality.
Note that the torus with a point removed has the free abelian group with generators $a,b$ as its fundamental group, whereas the torus has $\mathbb{Z} \oplus \mathbb{Z}$. If the induced map were injective, then we would have, by the isomorphism theorem, an isomorphism between a non-abelian group and a abelian group (the image of the induced map), which is a contradiction.

EDIT 2: I would like to point out two things. First, as stated in the comments, the rest of the answer given above applies to show that the torus minus a small open disk is not the retract of the torus, so it was not useless (it also shows a nice computation with the fundamental groups, and an example where using homology directly would fail).
Secondly, we don't need the full strength of compacity (using the fact that the image of continuous functions of compact sets are compact) for the result. In fact, any retract of a Hausdorff space must be closed.
This follows due to the fact that given a retract $r:X \to A$, we have that $A=f^{-1}(\Delta), $ where $f=r \times Id: X \times X \to A \times A$ and $\Delta$ is the diagonal. Since $X$ is Hausdorff, $A$ is Hausdorff, and also $A \times A$, therefore the diagonal $\Delta$ is closed. It follows that $A$ is closed, being the inverse image of a closed set by a continuous map. In the particular example of the question, $T \backslash \{x_0\}$ is not closed, a contradiction.
