# Disc with two points identified

Is a disc $D^2$ with two points on the boundary identified, same as $D^2 \vee D^2$ ? They both have boundary $S^1 \vee S^1$. I am confused because an exercise in Hatcher seems to ask the same question about the two spaces (i.e. to prove that there doesn't exist a retraction from each of this space to $S^1 \vee S^1$).

• Do you mean two points on the boundary? Commented May 13, 2013 at 20:06
• @Stefan: Yes, edited the post accordingly. Commented May 13, 2013 at 20:07
• They aren't even homotopy equivalent. The first one is homotopy equivalent to $S^1$ and the second is contractible. Commented May 13, 2013 at 20:14
• What happens if you take $D^2\vee D^2$ with the origin as the base point? Commented May 13, 2013 at 20:14

The two spaces are not even homotopy equivalent. I'll describe the intuition for the homotopy type of each space.

For $D^2$ with two (distinct) points on the boundary identified, visualize it as the cylinder $S^1 \times [0,1]$ with one part of the wall "pinched" to a point. It's easy to see this is homotopy equivalent to $S^1$.

$D^2 \vee D^2$ is contractible since we can just contract each copy of $D^2$ to the basepoint.

• Can you aid in visualising the homotopy between $S^1 \times [0,1]$ with one all pinched and $S^1$?
– user319128
Commented Aug 2, 2016 at 6:43
• @Jordan If $X = (S^1 \times [0,1])/\sim$ with $\sim$ being the equivalence relation that identifies all points of the form $(i,t)$, $t \in [0,1]$, then the deformation retract $H: (S^1 \times [0,1]) \times [0,1]$, $H(u,t,s) = (u, t(1-s))$ descends to a deformation retract from $X$ to the circle $\{[u, 0] : u \in S^1\} \subset X$. On $S^1 \times [0,1]$, $H$ is visualized as "pushing down" in the $[0,1]$-direction. Commented Aug 3, 2016 at 0:04
• May I ask another question? I just want to be clear. When we consider two distinct points to be identified, do we glue together to form a new surface? This is how I'm interpreting $D^2 \mapsto S^1 \times [0,1]$.
– user319128
Commented Aug 3, 2016 at 3:10

If you delete one point from the first space it is still connected. However if you delete the wedge point from $D^2 \vee D^2$, the resulting space is disconnected.