Proving Brouwer fixed point theorem in lower dimensions using vector calculus theorems By vector calculus theorems, I don't mean the general Stokes' theorem using differential forms. Rather I mean using Green's theorem for plane curves to prove Brouwer in dimension 2 or divergence theorem in dimension 3.
I assume it still boils down to deducing the non-existence of continuous retractions from $D^n$ to $S^{n-1}$. For $n=1$ (so $D^1=[-1,1]$ and $S^0=\{-1,1\})$ the argument seems easy. Assume there is a such a retraction $r$. $r$ must be constant (or else $r[-1,1]$ is disconnected), so $r'=0$. So the fundamental theorem of calculus yields
$$
0 = \int_{-1}^1 r'(t)dt = r(1)-r(-1)=1-(-1)=2,
$$
a contradiction.
I am essentially asking if there is an explicit application of the standard vector calculus theorems for $n=2,3$ in this spirit. Thanks for any help!
 A: I left this tab open for over a week before I finally came up with an answer.  Good question!
First, using the FTC in the $n=1$ case is a bit of overkill.  Like you said, a continuous function $r \colon [-1,1] \to \{-1,1\}$ would have to have a connected image, so it must be a constant.  But if $r$ is a retraction, then $r(-1) = -1$ and $r(1) =1$.  These can't both be true.  
As for $n=2$, suppose there is a $C^2$ retraction $F \colon D^2 \to S^1$.  Write $P$ and $Q$ for the components of $F$; since $F$ takes values in $S^1$ we have $P^2+Q^2 = 1$.  Differentiating with respect to $x$ and $y$ gives
\begin{align*}
    P \frac{\partial P}{\partial x} + Q \frac{\partial Q}{\partial x} &= 0 \\
    P \frac{\partial P}{\partial y} + Q \frac{\partial Q}{\partial y} &= 0 \\
\end{align*}
So
$$
    \begin{bmatrix} \frac{\partial P}{\partial x} & \frac{\partial Q}{\partial x} \\
    \frac{\partial P}{\partial y} & \frac{\partial Q}{\partial y} 
    \end{bmatrix}\begin{bmatrix} P \\ Q \end{bmatrix} = 0
$$
Since $(P,Q) \neq (0,0)$, we must have 
$$
    \begin{vmatrix}
        \frac{\partial P}{\partial x} & \frac{\partial Q}{\partial x} \\
        \frac{\partial P}{\partial y} & \frac{\partial Q}{\partial y} 
    \end{vmatrix}
    =\frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} 
     - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x} = 0
$$
Since $F$ is a retraction, we have $P(x,y) = x$ and $Q(x,y) = y$ for all $x,y \in S^1$.  This means that
$$
    \oint_{S^1} x\,dy = \oint_{S^1} P\,dQ
    = \oint_{S^1} P\left(\frac{\partial Q}{\partial x}\,dx + \frac{\partial Q}{\partial y}\,dy\right)
$$
Applying Green's theorem to the right-hand side gives
\begin{align*}
    \oint_{S^1} P\left(\frac{\partial Q}{\partial x}\,dx + \frac{\partial Q}{\partial y}\,dy\right)
    &= \iint_{D^2}\left[\frac{\partial}{\partial x}\left(P\frac{\partial Q}{\partial y}\right)- \frac{\partial}{\partial y}\left(P\frac{\partial Q}{\partial x}\right)
    \right]\,dA
    \\&= \iint_{D^2}\left[\left(\frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} + P \frac{\partial^2 Q}{\partial x\,\partial y}\right)- 
\left(\frac{\partial P}{\partial y}\frac{\partial Q}{\partial x}+P \frac{\partial^2 Q}{\partial y\,\partial x}\right)
    \right]\,dA
    \\&= \iint_{D^2}\left(\frac{\partial P}{\partial x}\frac{\partial Q}{\partial y} - \frac{\partial P}{\partial y}\frac{\partial Q}{\partial x}\right)\,dA = 0
\end{align*}
But applying Green's theorem to the left-hand side gives
\begin{align*}
    \oint_{S^1}x\,dy &= \iint_{D^2}\frac{\partial x}{\partial x}\,dA = \iint_{D^2}1 \,dA = \pi
\end{align*}
This is a contradiction. $\Box$

What's happening is a low-tech translation of the usual Stokes' theorem argument (from Wikipedia):
$$
0<\int_{S^{n-1}}\omega = \int_{S^{n-1}}F^*(\omega) = \int_{D^n}dF^*(\omega)= \int_{D^n}F^*(d\omega)=\int_{D^n}F^*(0) = 0
$$
where $\omega = x\,dy$ and $F^*\omega = P\,dQ$ (here's why).  Since the partial derivative matrix $\frac{\partial (P,Q)}{\partial (x,y)}$ has zero determinant, $d F^*\omega = dP \wedge dQ = 0$.
For $n=3$, I'm sure you can do something similar.  A function $F \colon D^3 \to S^2$, written in coordinates as $(P,Q,R)$, will satisfy $\left|\frac{\partial(P,Q,R)}{\partial(x,y,z)}\right| = 0$.  If $F$ is a retraction, then 
$$
    \iint_{S^2} x\,dy\,dz = \iint_{S^2} P\,dQ\,dR
$$
The LHS is nonzero by the divergence theorem.  The RHS should turn out to be
$$
    \iiint_{D^3} \left|\frac{\partial(P,Q,R)}{\partial(x,y,z)}\right|\,dV = 0
$$
