Consider the Euler sequence $$0\rightarrow \gamma \rightarrow \epsilon^4 \rightarrow Q \rightarrow 0$$ over $\mathbb{P}^3$. Take the projectivized bundle $\pi:P(Q)\rightarrow \mathbb{P}^3$ and consider its respective short exact sequence $$0\rightarrow L \rightarrow \pi^*Q \rightarrow Q' \rightarrow 0,$$ where $L$ is the tautological line bundle of $P(Q)$. Let $$t = c_1(L^*) \quad a = c_1(\gamma^*).$$ It is well - known by the Leray - Hirsch Theorem that the cohomology of $P(Q)$ is given by $$ H^*(P(Q)) \cong H^*(\mathbb{P}^3)[t]/(t^3 + at^2 + a^2t + a^3),$$ where the multiplication is defined as $\beta\cdot t^k = \pi^*\beta \cup t^k$ for $\beta \in H^*(\mathbb{P}^3) \cong \mathbb{Z}[a]/(a^4)$.
My question is the following: Denoting by $\pi^!: H^{2k}(P(Q))\rightarrow H^{2(k-2)}(\mathbb{P^3})$ the Gysin - or shriek - map of $\pi$, how do I show that $$\pi^!(t^2) = 1\quad \pi^!(t^3) = -a \quad \pi^!(t^4) = \pi^!(t^5) = 0?$$
What I have so far is that I chose numbers $\mu_0, \mu_1, \mu_2$ and $\mu_3$ such that $$\pi^!(t^2) = \mu_0\cdot 1\quad \pi^!(t^3) = \mu_1 \cdot a,\quad \pi^!(t^4) = \mu_2 \cdot a^2\quad \pi^!(t^5) = \mu_3 \cdot a^3$$ and then used the relation $t^3 + at^2 + a^2t + a^3 = 0$ to obtain equation the equation $$0 = \pi^!(t^3 + at^2 + a^2t + a^3) = (\mu_0 + \mu_1) a$$ and thus, I can easily obtain $$ 0 = \pi^!(t^4 + at^3 + a^2t^2 + a^3 t) = \pi^!(t^4) = \mu_2 a^2,$$ $$ 0 = \pi^!(t^5 + at^4 + a^2t^3 + a^3 t^2) = \pi^!(t^5) = \mu_3 a^3.$$ What I don't see is why $$\mu_0 = 1,\quad \mu_1= -1$$ since I only get $\mu_0 + \mu_1 = 0$. So there is one defining equation missing but I don't know where to get it from.
Thank you in advance for any reply!