# Showing that two spaces are homotopy equivalent

Let $x_0 \in S^1 \times S^1$. I want to show that $(S^1 \times S^1) - \{x_0\}$ and $S^1 \vee S^1$ are homotopy equivalent.

We have to show that $\exists$ maps $f: X \rightarrow Y$ and $g: Y \rightarrow X$ such that $g \circ f \simeq id_x$ and $f \circ g \simeq id_y$ (with $X = (S^1 \times S^1) - \{x_0\}$ and $Y = S^1 \vee S^1 = S^1 \amalg S^1$).

I want to find continuous functions $f$ and $g$ but cannot think of any. I was just wondering if somebody could give me a hint...

Think of $S^1 \times S^1$ as a square with opposite sides identified (this is precisely what you do when considering a torus), then remove a point inside it. It is now easy to construct a deformation retraction onto its boundary, which is $S^1 \vee S^1$.
• $S^1 \vee S^1$ is the pointed sum, as you claimed. But the map you're talking about is not a homeomorphism, it is just a homotopy equivalence: for example if you consider $S^1 \vee S^1 \setminus x$ (where $x$ is the common point) you get a disconnected space, while it is impossible to disconnect the torus minus a point by simply removing a point. – Edoardo Lanari Feb 13 '14 at 11:46
• I think you are confusing the concept of homotopy equivalence and that of isomorphism: try to find a map $X \to Y$ which is a homeomorphism (it is impossible, as I pointed out in the previous comment). On the contrary, you are just thinking to an homotopy from $Id_X$ to $i \circ r$, where $r$ is the above mentioned retraction onto $S^1 \vee S^1$. – Edoardo Lanari Feb 13 '14 at 11:54