Simply Connected domains. If $U$ and $U'$ be two domains in $\Bbb C$, and $f$ be a homeomorphism in $U$ and $U'$ then domain $U$ is simply connected $\iff$ $U'$ is simply connected. I found this problem in complex analysis. So I would prefer to know its proof from complex point of view rather using topological propositions. Thanks.
There are few properties which are equivalent for a domain $D$ in Complex plane. 
a)$D$ is simply connected.
b)for each $z_0\in \Bbb C$\ $D$ there is a analytic branch of $log(z-z_0)$ defined on $D$.
c)The compliment of $D$ in the extended complex plane $\Bbb C^*$ is connected.
 A: We will use the Riemann criterion for complex simply connected spaces.
Let $U$ be simply connected and $f$ be biholomorphic.
We have only to prove that if $g:U'\to \Bbb C$ holomorphic with $g(z)\neq 0$ for every $z$, then there is a holomorphic $g_1:U'\to \Bbb C:g(z)=g_1^2(z)$ for every $z$.
Let such a holomorphic function $g$.The composition $gof:U\to \Bbb C$ is holomorphic and $gof(w)\neq 0$ for every $w$. Because $U$ is simply connected ,there is a holomorphic $h:U\to \Bbb C:gof(w)=h^2(w)$ for every $w\in U$.
Let $hof^{-1}:U'\to \Bbb C$. Then it is holomorphic and for every $z\in U'$ we have that $(hof^{-1}(z))^2=(h(f^{-1}(z))^2=g(z)$ for every $z$.
A: What you say is true just by writing down the definition of homeomorphism and the definition of simply connected essentially.  Simply connected is a topological property, and homeomorphisms let you transfer topological properties between spaces.
What complex analysis does beautifully is to show that any two simply connected proper subsets of the plane are homeomorphic.  This is a really nontrivial result, and the Riemann mapping theorem solves it in a spectacular way:  by proving the much stronger result that they are biholomorphic!
A: The question is quite weird ; the notion of simply connectedness is a topological notion and the property you are asking is a lot simpler than complex analysis. If $U$ and $U'$ are homeomorphic , then any topological notion on $U$ can be carried to $U'$ by the homeomorphism.
For your question, this can take the following form. Assume that $U$ is simply connected. Choose a point $y$ in $U'$ and take a loop $\gamma(t)$ with base-point $y$ in $U'$. Then $f^{-1} \circ \gamma(t)$ is a loop in $U$ with base-point $x = f^{-1}(y)$. Since $U$ is simply connected, you have an homotopy $F(s,t)$ from $f^{-1} \circ \gamma$ to the constant loop based at $x$. Then $f \circ F(s,t)$ is a homotopy from $\gamma$ to the constant loop based at $y$.
A: It seems to me like an incredible overkill to use the Riemann mapping theorem here. Perhaps you can use the following theorem:

Let $U$ be a connected open subset of $\mathbf C$. Then $U$ is simply-connected if and only if for every holomorphic function $g$ on $U$ and every closed path $C$ in $U$, $\int_C g(z) dz = 0$.

Then, under the assumption that $U$ is simply connected, use $f$ and the change-of-variables formula to prove that the same criterion holds in $U'$.
(Of course, the real reason is a purely topological one: two homeomorphic spaces have isomorphic fundamental groups. I'm only giving this weird solution because you insist on avoiding "topological" arguments.)
