Prove $\mathbb{D}$ and $\mathbb{D} \cup \{(0,1)\}$ aren't homeomorphic. They give me $\mathbb{D}$ = {$(x,y) \in \mathbb{R^2}: x^2 + y^2 < 1$} and they ask me to prove that $\mathbb{D}$ and $\mathbb{D} \cup \{(0,1)\}$ aren't homeomorphic.
My idea to prove this is using $f \colon \mathbb{D} \cup \{(0,1)\} \to \mathbb{D}$ and taking the fundamental group of $\mathbb{D}$ and $\mathbb{D}\setminus\{f(0,1)\}$ in one random point, so the first one has the trivial set as fundamental group and the second has something different, so as the fundamental group is a homeomorphism invariant they aren't homeomorphic.
I don't know if it is ok or not.
 A: $\mathbb D$ is a $2$-manifold: $\mathbb D \cup\{(0,1)\}$ is not, as no neighbourhood of $(0,1)$ in it is homeomorphic to $\mathbb R^2$.  For example, no neighbourhood of $(0,1)$ is compact. 
A: I think you have the right idea, but you need to be more careful about how you phrase your argument. The key point is that every point $x\in\mathbb{D}$ has the property that $\mathbb{D} \setminus x \simeq S^1$, but $(\mathbb{D}\cup \{(0,1)\}) \setminus (0,1) = \mathbb{D} \simeq *$. If there were a homeomorphism $f\colon \mathbb{D}\cup \{(0,1)\} \to \mathbb{D}$ then we would have to have both $S^1 \simeq \mathbb{D}\setminus f(0,1) \simeq *$, which is a contradiction since $S^1$ is not contractible.
A: Name $P=(0,1)$ and  $\mathbb D^\prime = \mathbb D \cup \{P\}$.
If $\mathbb D$ and $\mathbb D^\prime$ would be homeomorphic, $\mathbb D =\mathbb D^\prime \setminus \{P \}$ would be homeomorphic to $\mathbb D $ minus a point. But that can’t be as $\mathbb D$ minus a point is not simply connected while $\mathbb D$ is.
