# 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.

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.

• Thanks! How do we know the torus with a point removed is not compact? – Mark Apr 30 '16 at 4:45
• @Mark One way to show this is using the fact that $T$ is Hausdorff and connected. It $T \backslash \{x_0\}$ were compact, it would be closed. But it is open, since it is the complement of a point. Since $T$ is connected, then $T \backslash \{x_0\}$ must not be closed. You can also argue directly that it will not be closed, since $x_0$ is clearly a limit point of its complement. Another way is that the torus is a compact metric space, hence complete. If you remove one point, it is not complete anymore (since such a point was clearly not isolated). Therefore, it cannot be compact. – Aloizio Macedo Apr 30 '16 at 4:58
• Sorry, for stupid question. But isn't the fundamental group of torus without a point free non-abelian? Why is it abelian? – Mihail Apr 30 '16 at 6:12
• I think your original proof was better. I think that a compact set (e.g., the origin) can be a retract of a noncompact set (e.g., the unit open disk), indeed a deformation retract, the retraction in this example being $H(x, t) = (1-t)x$. But even if I've got this wrong, the remainder of your proof applied to the torus minus an open disk about $x_0$, which is equally interesting. :) – John Hughes Apr 30 '16 at 13:04
• @JohnHughes A compact set can be a retract of a noncompact set, but not conversely, since continuity preserves compactness (and the question is about the converse situation). But you are right, the remained of the proof applies to the torus minus an open disk : ). – Aloizio Macedo Apr 30 '16 at 21:28