Sketching the image of the function $\hat{h}.$ I am trying to solve a puzzle through solving a series of questions, here is the first version of the  puzzle:
Is it possible to find 2 connected subsets $A$ and $B$ inside the square $I \times I$ such that $(0,0),(1,1) \in A$ and $(0,1),(1,0) \in B$ and $A \cap B = \emptyset.$
We know that the answer is YES. But we are guessing that the answer is NO if we required $A$ and $B$ to be path connected.Our intuition comes from the attitude of the Topologist's sine curve. Now we are trying to write a proof of this guess. During this path we wanted to prove the Jordan Curve Theorem. So I am trying to do so.
Here are the questions that I should have solved so far:
Proving the equivalence of the Criss - Cross theorem statement(ordinary one) to another statement.
Proving the criss cross theorem (generalized version). (I did not solve this question yet, so any help with the solution will be appreciated )
Also I know the proof of the ordinary version of the criss-cross theorem. Here is it:
Assume towards contradiction that we have such paths $\alpha$ and $\beta$ such that $\alpha (s) \neq \beta (t)$ for all $s,t \in I.$ This can be written as $\alpha (s)- \beta (t)  \neq 0$ for all $s,t \in I.$ Which can further be rephrased by defining $H: I \times I \rightarrow \mathbb{R}^2$ by the formula $$H(s,t) = \alpha(s) - \beta (t).$$ And our assumption means that the function $H$ never takes the value $0 \in  \mathbb{R}^2;$ in diagram language, we assume that there is a continuous function $\hat{H}$ making the following diagram commutative:

$$\require{AMScd}
\begin{CD}
I \times I @>{\hat{H}}>> \mathbb{R}^2 - \{0\}\\
@VVV @VVV \\
I\times I @>{H}>> \mathbb{R}^2
\end{CD}$$
where the arrow below $\hat{H}$ should be a dotted arrow because we are searching for this function. And I am not skillful in drawing commutative diagrams this is why I draw $I \times I$ 2 times because I do not know how to draw one dotted arrow coming out of $I \times I$ going directly to $\mathbb{R} - \{0\}$ my bad. Then my job is to show that there can be no such function $\hat{H}.$ 

Now, I want to solve this problem:
Suppose there were such a function $\hat{H},$ and write $\hat{h}$ for the restriction of $\hat{H}$ to the boundary of $I \times I.$ Sketch what the image of $\hat{h}$ might look like - remember that there is no guarantee or assumption that either $\hat{H}$ or $\hat{h}$ is injective, so put some possible self-crossing in your sketch.
I do not know how to sketch it. could anyone help me in doing so please?
 A: The key observation is that you know a little bit about what $\hat{h}$ looks like.
Assume $\alpha : (0,0)\to (1,1)$ and $\beta : (0,1)\to (1,0)$ are your paths.
Then
$$\hat{h}(0,0) = \alpha(0)-\beta(0) = (0,0)-(0,1) = (0,-1).$$
as $t$ runs from $0$ to $1$, we get
$$\hat{h}(t,0) = \alpha(t)-\beta(0) = (q,r) - (0,1) = (q,r-1).$$
We know almost nothing about this path except that 
$0\le q\le 1$ and $-1\le r\le 0$ and $(q,r)\ne (0,1)$, so drawing any path that stays inside this square and doesn't touch the origin is acceptable.
Then $\hat{h}(1,0) = (1,1)-(0,1) = (1,0)$.
For $0\le s\le 1$, 
$\hat{h}(1,s)$ stays in $[0,1]\times[0,1]\setminus\{(0,0)\}.$
Again, drawing any path that stays in here is acceptable.
Next
$\hat{h}(1,1) = (1,1)-(1,0) = (0,1)$, and as 
$t$ goes from $1$ back to $0$ on the next segment, $\hat{h}(t,1)$
stays inside $[-1,0]\times [0,1]$ and doesn't touch the origin.
Finally, $\hat{h}(0,1) = (-1,0)$, and as $s$ returns to $0$, $\hat{h}(0,s)$
stays inside $[-1,0]\times[-1,0]$ and doesn't touch the origin before returning to $(0,-1)$.
You'll notice that any path you try to draw that satisfies these requirements forms 
a counterclockwise loop about the origin that is nonzero in the fundamental group, contradicting the assumption of the existence of the extension $\hat{H}$. 
Note that this proves the sets in your puzzle cannot exist if they are path connected. Not sure why you would need the Jordan curve theorem.
