Is there a continuous surjective map from a 2-cell to a 2-sphere? I think there is, tho there probably isn't a differentiable one. My idea is to consider squares centered at origin in $[-1,1]^2$ and send them to circles cut by a plane in $S_2:= \{p \in \mathbb{R}^3 : |p| = 1\}$. (The origin point would go to the point when the plane intersects the sphere at only 1 point, and the outer square would go the the other one). Hopefully if the way the squares are sent is "consistent" (so that near squares don't for example send points in same face of square to opposite sides of the circle), then gluing this all together we'd have the desired map. But i have trouble formalizing and proving whether this works...
 A: Ok, here's a very step-by-step proof.
I assume your definition of $2$-cell is the convex hull $H$ of the points $(0,0), (1,0)$ and $(0,1)$ in the plane.  Inscribe a circle $C$ in $H$ and let $x$ be the center of the disc inside $C$.  Let $t_h$ be the line segment from $x$ to $h$ for each $h$ in the boundary of $H$, denoted $\partial(H)$.  Then since $H$ is convex, $t_h$ intersects $\partial(H)$ in a single point and is contained in $H$.
If $y$ is in $H$ then there is some $t_h$ containing $y$ by convexity, and since these segments only intersect in the point $x$, each point in $H$ but outside of $C$ is contained in exactly one such $t_h$.  For each $y$, let $b(y)$ be the unique point on $C$ sharing such a segment containing $y$.
Define a function $f$ on $H$ that's equal to the identity inside of $C$ and its boundary, and otherwise is equal to $b(y)$.  It's surjective by definition, and continuous on $C$ and the disc $D$ inside it, since it's just the identity there.  We just need to show it's continuous outside of $\overline{D}$.  Let $y$ be such a point and let $y_n \rightarrow y$.
Then by convexity there are line segments $s_n$ between $y_n$ and $y$ respectively, whose lengths converge to zero since $y_n \rightarrow y$.  Thus the angle $\alpha_n$ of the triangle between $x, y$ and $y_n$ at the vertex $x$ converges to zero.  If $R_n$ is one of the other sides of this triangle, then $\alpha_n \rightarrow 0$ implies that $R_n \cap C \rightarrow t_y \cap C = b(y)$.  But that's the definition of $f$, and this shows that it's continuous since these are compact metric spaces.
This shows that the $2$-cell has a continuous surjection onto the closed disc.  Then the one-point compactification of the open disc is the sphere, and this extends continuously to a function on the closed disc by mapping $\partial(D)$ to $\infty$.  Composition gives your function.
Smooth surjections do exist, as well, but the details involved to construct it explicitly are more complicated.  A sketch: Take an open disc in the triangle away from the boundary and do the one-point compactification there as well, extending it beyond the boundary to the rest of the $2$-cell by having the function 'continue' along the $t_h$'s which will be mapped to (subarcs of) great circles through infinity.  Basically 'overwrap' the sphere with the plane so that your $2$-cell spills over $\infty$.
