Homeomorphism with a bouquet of two circles I understand why removing two points from $\mathbb R^2$ gives a surface that is homeomorphic to a bouquet of two circles. But can someone please write this homeomorphism?
 A: As it has been remarked in the comments, $\mathbb{R}^2\backslash\{p,q\}$ is not homeomorphic to a bouquet of two circles! However there is a deformation retraction of that space to a subspace homeomorphic to the bouquet of two circles, in particular they have the same homotopy type. The deformation retraction can be written down explicitely, but it is a bit messy and not really enlightening. If you want it, write a comment and I will write it down.

Ok, I will explain how I would construct a deformation retraction from $X=\mathbb{R}^2\backslash\{-1,1\}$ to a subspace $Y\subset X$ homeomorphic to $S^1\vee S^1$. Notice that the retraction induced this way is a homotopy equivalence between the two spaces (with inverse the inclusion $Y\to X$).
Let $Z$ be the union of the circle of radius $2$ centered in $0$ with the vertical segment from $(0,-1)$ to $(0,1)$. We define a deformation retraction $F:[0,1]\times X\to X$ by:
$$F(t,x)=\cases{\left(\frac{t}{\|x\|}+(1-t)\right)x & if $\|x\|\ge2$\\
tz(x)+(1-t)x&else}$$
where $z(x)$ is $x$ if $x_1=0$, and the intersection which is closest to $x$ of the straight  line passing through $x$ and the closest point to $x$ between $(0,-1)$ and $(1,0)$ (you can write this down explicitly, but you have to separate a couple more cases). This map deforms $X$ to $Z$ keeping $Z$ fixed.
Next define $G:[0,1]\times Z\to X$ by $G(t,x)=ty(x)+(1-t)x$, where $y(t)$ is the point of $Y$ which is closest to $x$ along the vertical line passing through $x$ (again, you can write this down easily if you want). This maps deforms the space $Z$ to $Y$.
The map $H:[0,1]\times X\to X$ defined by:
$$H(x,t)=\cases{F(2t,x)&if $t\le1$\\G(2t-1,F(1,x))&if $t\ge1$}$$
is the desired deformation retraction.
