Vanishing cycles exact sequence for degeneration of curves Let $f : X\rightarrow D$ be a family of smooth projective curves over the complex unit disk $D$, degenerating to a nodal curve $X_0$ above $0\in D$.
Let $\eta\in D - \{0\}$ be a general point, and let $\gamma\subset X_\eta$ be the vanishing cycle, which for me is a disjoint union of simple closed curves. Let $T\supset\gamma$ denote a tubular neighborhood of $\gamma$ in $X_\eta$, so $T$ is a disjoint union of annuli.
I believe there should be an exact sequence (in betti cohomology)
$$0\rightarrow H^1(X_0)\rightarrow H^1(X_\eta)\rightarrow H^1(T)\stackrel{\delta}{\rightarrow }H^2(X_0)\rightarrow H^2(X_\eta)\rightarrow H^2(T) = 0$$
The map $H^*(X_0)\rightarrow H^*(X_\eta)$ comes from the inclusion $X_\eta\rightarrow X$ using the isomorphism $H^*(X) = H^*(X_0)$.

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*Is there a geometric description of the connecting map $\delta$?

*Does anyone have a good reference for this exact sequence?

*Am I correct in saying that this sequence is equivariant for the monodromy action of $\pi_1(D-\{0\},\eta)$, where $H^1(X_\eta)$ is the only term where the action is nontrivial?

To set some context, while I'd eventually want a deeper understanding of vanishing cycles, I'd like to begin by understanding this simple example, and it's hard to find a reference that doesn't immediately dive into perverse sheaves and such.
 A: This is from standard Lefschetz's theory, which works for a one-parameter family with special fiber acquiring a node. The fiber can have arbitrary dimensions, not only for curves.
To start with, I don't see where the $H^1(T)$ in your sequence is coming from. To me, there is a long exact sequence of the cohomology of the pair $(X,X_{\eta})$:
$$\cdots \to H^1(X)\to H^1(X_{\eta})\xrightarrow{\alpha} H^2(X,X_{\eta})\to H^2(X)\to \cdots.\tag{1}\label{1}$$
First $X$ deformation retracts onto $X_0$, so one has $H^*(X)\cong H^*(X_0)$. Second, by excision, $H^*(X,X_{\eta})\cong H^*(U,U\cap X_{\eta})$, where $U$ is an open ball in $X$ around the nodal point on $X_0$. The pair $(U,U\cap X_{\eta})$ is homotopic equivalent to $(C\gamma,\gamma)\simeq (B^2,S^1)$, where $C\gamma$ is the cone over the vanishing cycle $\gamma$. So $H^2(X,X_{\eta})\cong H^2(B^2,S^1)$ is generated by the single class $[C\gamma]$. Since $\alpha$ is not zero map, we conclude that the long exact sequence \eqref{1} reduces to
$$0\to H^1(X_0)\to H^1(X_{\eta})\xrightarrow{\alpha} \mathbb Z\to 0, \ \text{and} $$
$$0\to H^2(X_0)\to H^2(X_{\eta})\to 0.$$
$\alpha$ sends the vanishing cycle to its cone, and $\ker(\alpha)$ is called the module of invariant cycles.
Sometimes, people work for $X$ projective and the base being $\mathbb P^1$, e.g., a Lefschetz pencil. In that case, $\alpha$ in \eqref{1} is still surjective, but there can be many vanishing cycles.
One can refer to [Voisin, vol II, Chapter 2 and 3], [Nicolaescu's notes], and [Lamotke, 81],
