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Let me write $x_i$ and $y_i$ for a symplectic basis of cohomology of $C$, and $a_i$, $b_i$ for the linear dual basis of the first homology of $C$. It is enough to find $\delta^*(dx_i)$ and $\delta^*(dy_i)$ in terms of $x_i$ and $y_i$.

But to do this we just evaluate $\langle \delta^*(dx_i), a_i \otimes 1 \rangle$, $\langle \delta^*(dx_i), b_i \otimes 1 \rangle$ and so on, which is easy as $\delta_*(a_i \otimes 1) = a_i$, $\delta_*(1 \otimes a_i) = -a_i$ and so on.

Hence $\delta^*(dx_i) = x_i \otimes 1 - 1 \otimes x_i$ and $\delta^*(dy_i) = y_i \otimes 1 - 1 \otimes y_i$. Hence $$\delta^*(\phi) = \sum (x_i \otimes 1 - 1 \otimes x_i)(y_i \otimes 1 - 1 \otimes y_i)$$ and multiplying the whole mess out and using relations like $x_iy_i = [C] = \tfrac{1}{\chi(C)} \psi$ (which perhaps has a sign, depending on your conventions), you get the required expression.


Addendum: You may also be interested in the following. In "Relations among tautological classes revisited" I gave the following alternative description of the class $\delta^*(\phi)$ on $\mathcal{C}_g^2$, the square of the universal curve, which parametrises curves with two ordered non-necessarily distinct points. The paper uses different notation, but I will stick to yours.

Let $\pi : \overline{\mathcal{C}}_g^2 \to \mathcal{C}_g^2$ be the universal curve, and $\Delta_i \in H^2(\overline{\mathcal{C}}_g^2)$ for $i=1,2$ denote the class of the locus of the first and second points respectively. Then $$\delta^*(\phi) = -\pi_!((\Delta_1 - \Delta_2)^2).$$ Again, depending on your conventions there may be a sign and a power of 2 somewhere.

Let me write $x_i$ and $y_i$ for a symplectic basis of cohomology of $C$, and $a_i$, $b_i$ for the linear dual basis of the first homology of $C$. It is enough to find $\delta^*(dx_i)$ and $\delta^*(dy_i)$ in terms of $x_i$ and $y_i$.

But to do this we just evaluate $\langle \delta^*(dx_i), a_i \otimes 1 \rangle$, $\langle \delta^*(dx_i), b_i \otimes 1 \rangle$ and so on, which is easy as $\delta_*(a_i \otimes 1) = a_i$, $\delta_*(1 \otimes a_i) = -a_i$ and so on.

Hence $\delta^*(dx_i) = x_i \otimes 1 - 1 \otimes x_i$ and $\delta^*(dy_i) = y_i \otimes 1 - 1 \otimes y_i$. Hence $$\delta^*(\phi) = \sum (x_i \otimes 1 - 1 \otimes x_i)(y_i \otimes 1 - 1 \otimes y_i)$$ and multiplying the whole mess out and using relations like $x_iy_i = [C] = \tfrac{1}{\chi(C)} \psi$ (which perhaps has a sign, depending on your conventions), you get the required expression.

Let me write $x_i$ and $y_i$ for a symplectic basis of cohomology of $C$, and $a_i$, $b_i$ for the linear dual basis of the first homology of $C$. It is enough to find $\delta^*(dx_i)$ and $\delta^*(dy_i)$ in terms of $x_i$ and $y_i$.

But to do this we just evaluate $\langle \delta^*(dx_i), a_i \otimes 1 \rangle$, $\langle \delta^*(dx_i), b_i \otimes 1 \rangle$ and so on, which is easy as $\delta_*(a_i \otimes 1) = a_i$, $\delta_*(1 \otimes a_i) = -a_i$ and so on.

Hence $\delta^*(dx_i) = x_i \otimes 1 - 1 \otimes x_i$ and $\delta^*(dy_i) = y_i \otimes 1 - 1 \otimes y_i$. Hence $$\delta^*(\phi) = \sum (x_i \otimes 1 - 1 \otimes x_i)(y_i \otimes 1 - 1 \otimes y_i)$$ and multiplying the whole mess out and using relations like $x_iy_i = [C] = \tfrac{1}{\chi(C)} \psi$ (which perhaps has a sign, depending on your conventions), you get the required expression.


Addendum: You may also be interested in the following. In "Relations among tautological classes revisited" I gave the following alternative description of the class $\delta^*(\phi)$ on $\mathcal{C}_g^2$, the square of the universal curve, which parametrises curves with two ordered non-necessarily distinct points. The paper uses different notation, but I will stick to yours.

Let $\pi : \overline{\mathcal{C}}_g^2 \to \mathcal{C}_g^2$ be the universal curve, and $\Delta_i \in H^2(\overline{\mathcal{C}}_g^2)$ for $i=1,2$ denote the class of the locus of the first and second points respectively. Then $$\delta^*(\phi) = -\pi_!((\Delta_1 - \Delta_2)^2).$$ Again, depending on your conventions there may be a sign and a power of 2 somewhere.

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Let me write $x_i$ and $y_i$ for a symplectic basis of cohomology of $C$, and $a_i$, $b_i$ for the linear dual basis of the first homology of $C$. It is enough to find $\delta^*(dx_i)$ and $\delta^*(dy_i)$ in terms of $x_i$ and $y_i$.

But to do this we just evaluate $\langle \delta^*(dx_i), a_i \otimes 1 \rangle$, $\langle \delta^*(dx_i), b_i \otimes 1 \rangle$ and so on, which is easy as $\delta_*(a_i \otimes 1) = a_i$, $\delta_*(1 \otimes a_i) = -a_i$ and so on.

Hence $\delta^*(dx_i) = x_i \otimes 1 - 1 \otimes x_i$ and $\delta^*(dy_i) = y_i \otimes 1 - 1 \otimes y_i$. Hence $$\delta^*(\phi) = \sum (x_i \otimes 1 - 1 \otimes x_i)(y_i \otimes 1 - 1 \otimes y_i)$$ and multiplying the whole mess out and using relations like $x_iy_i = [C] = \tfrac{1}{\chi(C)} \psi$ (which perhaps has a sign, depending on your conventions), you get the required expression.