As the title says, I want to embed the genus 4 surface inside $\mathbb{C}P^2\# \mathbb{C}P^2$ representing a nontrivial homology class.

I know that $H^2(\mathbb{C}P^2 \# \mathbb{C}P^2; \mathbb{Z})\simeq \mathbb{Z} \oplus \mathbb{Z}$ where generators come from generators for $H^2$ of each copy of $\mathbb{C}P^2$, and these can be taken to be represented by the homology class of a hyperplane (degree 1 algebraic curve), denoted, say, by $[H]$. So my first attempt was finding some way of embedding the genus 2 surface inside $\mathbb{C}P^2$, then take a two disk in this surface, multiply it by another two disk to get a 4 disk inside $\mathbb{C}P^2$, and this is the 4 disk I would remove to make the connect sum $\mathbb{C}P^2\# \mathbb{C}P^2$. If I could manage this so that the genus 2 surface is an algebraic curve in $\mathbb{C}P^2$ of degree $d$, say, then it would represent $d[H]$ (not entirely sure about this also, though), and hence the connect sum would represent $(d,d)$ in $H^2(\mathbb{C}P^2 \# \mathbb{C}P^2; \mathbb{Z})$. But I read somewhere (I didn't know this but I guess is rather standard for people who know about this things; not my case!), that if $g$ is the genus of an algebraic curve in $\mathbb{C}P^2$ of degree $d$, then we get $g=(d-1)(d-2)/2$, so no good for $g=2$.

So there goes a failed attempt. Appreciate any help.

As the title says, I want to embed the genus 4 surface inside $\mathbb{C}P^2\# \mathbb{C}P^2$ representing a nontrivial homology class.

I know that $H_2(\mathbb{C}P^2 \# \mathbb{C}P^2; \mathbb{Z})\simeq \mathbb{Z} \oplus \mathbb{Z}$ where generators come from generators for $H^2$ of each copy of $\mathbb{C}P^2$, and these can be taken to be represented by the homology class of a hyperplane (degree 1 algebraic curve), denoted, say, by $[H]$. So my first attempt was finding some way of embedding the genus 2 surface inside $\mathbb{C}P^2$, then take a two disk in this surface, multiply it by another two disk to get a 4 disk inside $\mathbb{C}P^2$, and this is the 4 disk I would remove to make the connect sum $\mathbb{C}P^2\# \mathbb{C}P^2$. If I could manage this so that the genus 2 surface is an algebraic curve in $\mathbb{C}P^2$ of degree $d$, say, then it would represent $d[H]$ (not entirely sure about this also, though), and hence the connect sum would represent $(d,d)$ in $H_2(\mathbb{C}P^2 \# \mathbb{C}P^2; \mathbb{Z})$. But I read somewhere (I didn't know this but I guess is rather standard for people who know about this things; not my case!), that if $g$ is the genus of an algebraic curve in $\mathbb{C}P^2$ of degree $d$, then we get $g=(d-1)(d-2)/2$, so no good for $g=2$.

So there goes a failed attempt. Appreciate any help.