Consider the function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ given by $f(x,y)=(x^2+x-2y^2+1,-x^2+y^2+3y-2)$. I am trying to show that the graph of this function is transversal to the diagonal $\Delta=\{(x,y,x,y)\}$ in $\mathbb{R}^2\times\mathbb{R}^2$. To do this, I want to show that the sum of the tangent spaces of each manifold at every point in their intersection equals $\mathbb{R}^2\times\mathbb{R}^2$.
Now, the tangent space of $f$ is the image of the Jacobian of $f$.The Jacobian of $f$ is given by $df=\begin{bmatrix} 2x+1 & -4y \\ -2x & 2y+3 \end{bmatrix}$. The tangent space of the graph of $f$ is the image of this map. It appears that $df$ will always be full rank regardless of the choices for $x$ and $y$.
However, I am confused about what the intersection should look like and how to make sure the sum of the tangent spaces is as desired. What should I do?
The intersection of the graph of $f$ (I'll denote this by $G(f)$) and $\Delta$ is the set of all points in the diagonal $(x,y,x,y)$ such that $$x=x^2+x-2y^2+1,\qquad y=-x^2+y^2+3y-2.$$ Solving this, you obtain $$G(f)\cap\Delta=\{(1,1,1,1),(-1,1,-1,1)\}.$$ From here it is just linear algebra: We want to show that for each $x\in G(f)\cap\Delta$, the map $T_xG(f)\oplus T_x\Delta\to T_x\mathbb R^4$, $(u,v)\mapsto u+v$, has full rank. The tangent space of $G(f)$ at $(x,f(x))$ is the set of all vectors $(u,df_xu)$, where $u\in T_x\mathbb R^2$, and the tangent space of $\Delta$ at $(x,x)$ is the set of all vectors $(v,v)$, where $v\in T_x\mathbb R^2$.
Thus, we need to show that the matrices $$\begin{pmatrix}I_2& I_2\\df_{(1,1)}&I_2 \end{pmatrix}$$ and $$\begin{pmatrix}I_2& I_2\\df_{(-1,1)}&I_2 \end{pmatrix}$$ are full rank, where $I_2$ is the $2\times 2$ identity matrix.