TITLE: Degree of map obtained by intersecting tangents to complex projective curve with fixed line
QUESTION [6 upvotes]: Let $X \subset \mathbb{C}P^2$ be a smooth projective curve, and for each $p \in X$, let $T_p$ denote the line tangent to $X$ at $p$.
For any line $L \subset \mathbb{C}P^2$ in the plane, there is a map $X \to L$ that sends $p \in X$ to the unique point of intersection $T_p \cap L \in L$. 
How does the topological degree of this map vary with $L$?

REPLY [4 votes]: The tangent map $f$ has natural range space $(\mathbb{CP}^2)^*$ the $\mathbb{CP}^2$ of lines in the original $\mathbb{CP}^2$. The degree of the image curve of $f$ is $d(d-1)$ where $d$ is the degree of the original curve. When you ask for the point of interaction with a fixed line $L$, you are considering the mapping from$$(\mathbb{CP}^2)^* \to L$$defined by intersection. Per the discussion, this mapping is not defined at the point $\{L\}$ of $(\mathbb{CP}^2)^*$ corresponding to $L$ itself. However, it becomes a well-defined map$$g: B \to L$$from the blowup $B$ of $(\mathbb{CP}^2)^*$ at the point $\{L\}$. So the degree of the composition $g \circ f$ can indeed change if $\{L\}$ is in the image of $f$. It goes down according to the multiplicity of intersection of the image of $f$ with the $\mathbb{CP}^1$ in $B$ that is collapsed to $\{L\}$ in $(\mathbb{CP}^2)^*$.