$\mathbb{C}^2\setminus \{(z,z)\ :\ z\in \mathbb{C}\}$ is path connected. I am really surprised that why $$ \mathbb{C}^2\setminus \{(z,z)\ :\ z\in \mathbb{C}\} $$ is path connected?
I was thinking that it is same as $\mathbb{R}^4\setminus\mathbb{R}^2.$ But still I am unable to prove that the above space is path connected. 
 A: Say you want a path from $(a,b)$ to $(c,d)$ within $\mathbb{C}^2\setminus\mathrm{diag}(\mathbb{C})$.
Here are a couple ways to think about this. Within $\mathbb{C}$, you'll be able to find paths $a\to c$ and $b\to d$, which you can combine in $\mathbb{C}^2$ by either traversing at the same time (componentwise) or one after the other. Within the plane, you can traverse straight line segments between two points or, if you're trying to avoid a point on said line segment, around a semicircle containing the two points.
See if you can get an answer with these hints.
A: Like how $\Bbb R^3$ removing a line (an affine subspace of two lower dimension) is path-connected, $\Bbb R^4$ removing a plane (also an affine subspace of two lower dimension) should, by guessing, also be path-connected.
First have a look how $\Bbb R^3$ removing a line is path-connected. Consider $\Bbb R^3\setminus L$, where $L=\{(0,0,x_3)\}$. Consider the cylinder $C=\{(x_1,x_2,x_3):x_1^2+x_2^2=1\}$. The cylinder is path-connected, which you can actually see. Each point $(x_1,x_2,x_3)$ on $\Bbb R^3\setminus L$ can be connected to $C$ by a straight line path that goes in the direction $(x_1,x_2,0)$ or its reverse direction. So the whole $\Bbb R^3\setminus L$ is path-connected.
For your case, the set would be $\Bbb R^4\setminus P$ with $P=\{(0,0,x_3,x_4)\}$. You again prove that the cylinder $C=\{(x_1,x_2,x_3,x_4):x_1^2+x_2^2=1\}$ is path-connected, and any point $(x_1,x_2,x_3,x_4)$ on $\Bbb R^4\setminus P$ can be connected to the cylinder by the straight line path that goes in the direction $(x_1,x_2,0,0)$.
