Deformation retraction of closed disk on open disk I'm wondering whether there exists a (strong) deformation retraction from the unit disk $D^2$ to its interior $B^2$.
My intuition says no but I couldn't come up with a proof, also because $B^2$ and $D^2$ are contractible, hence homotopic equivalent. Do you have any hint on how to proceed?
 A: The image of a retraction is always closed in that space (if the space is Hausdorff ) and the interior $B^2$ is not closed in $D^2$. 
A proof using sequences: if $r: X \to A$ is a retraction, and $(a_n)_n$ in $A$ converges in $X$ to $x$, then $r(a_n) \to r(x)$ by continuity of $r$. Being a retraction, $r(a_n)=a_n$ and as limits are unique, $r(x)=x$ so that $x \in A$, and so $A$ is closed.
A: If $A \subset X$ is a dense subest with inclusion map $i$ then if $Y$ is Hausdorff if for two continuous functions $f,g:X \rightarrow Y$, $f \circ i = g \circ i$ then $f = g$. 
I'm gonna use this to prove that $B^2$ is not a retract of $D^2$ by contradiction. 
If $r:D^2 \rightarrow B^2$  is retract and $i:B^2 \rightarrow 
D^2$ is the inclusion then $r \circ i = id_{B^2}$ by the definition of retract so $i\circ r \circ i = id_{B^2} \circ i = id_{D^2} \circ i$. 
Set $i \circ r = f$ and $g = id_{D^2}$ then this means that $f \circ i = g \circ i$ so it must be the case that (since $D^2$ is Hausdorff and $B^2$ is dense in $D^2$) $i \circ r = id_{D^2}$ so $i$ and $r$ are homeomorphisms which are inverses to eachother, obviously this is not the case since $i$ is not even a bijection.
The existence of a retraction is weaker than the existence of a deformation retract which is weaker than the existence of a strong deformation retract. So a strong deformation retract implies the existence of a retract. The result follows.
A: Yet another proof for the negative answer: Continuous images of compact spaces are compact. So there's no continuous map of the closed disk onto its interior.
