I am presented with the following problem:
Given $\pi: S^3 \to \mathbb{R}P^3$, the quotient map to the real projective space.
For any $t \in [0,1]$, we define $S_t=\{(x,y,z,t): x^2+y^2+z^2+t^2=1\} \subset S^3$
And we define $\Lambda:= \{(x,0,0,w): x^2+y^2=1\}\subset S^3$.
Prove that for any $t \in [0,1]$: $\pi(S_t)$ is a dimension $2$ sub-manifold of $\mathbb{R}P^3$, that intersects $\pi(\Lambda)$ transversally.
Ok.. so, proving that $\pi(S_t)$ is a sooth manifold for $t>0$, is pretty trivial, considering that $S_t$ is a smooth manifold, and $S_t$ is homeomorphic to $\pi(S_t)$.
And when $t=0$, just pick local charts that sends an small enough open set back to $S_t$, to specified, strategically chosen open sets, then sending them by the charts of $S_t$.
For the ladder part, I thought of this, since $Y:=\pi(\Lambda)$ has dimension $1$ and $Z:=\pi(S_t)$ dimension $2$, you have that given $x\in Y\cap Z$,
Since $dim(T_xY\cap T_xZ)$ can be at most $1$:
$dimT_xY + dimT_xZ-dim(T_xY\cap T_xZ)$ is either $3$ or $2$. (If it is $3$ I am done)
My idea is to prove that $T_xY\cap T_xZ$ cannot have dimension $1$, so if $(T_xY\cap T_xZ) \subset T_xY$ has dimension $1$, you must have $(T_xY\cap T_xZ) = T_xY$, hence $T_xY \subset T_xZ$.
Now, all I got to do is find an $v \in T_xY$ that it is not in $T_xZ$, but my problem is seeing $T_xY$ and $T_xZ$. I tried proving that $\pi$ is differentiable submersion but I am not sure if this is true.
Any help would be appreciated.