Degree of map $S^2\times S^2$ to $\mathbb{C}P^2\# \mathbb{C}P^2$ I want to know which are the possible degrees of a map $f\colon S^2\times S^2 \longrightarrow \mathbb{C}P^2 \# \mathbb{C}P^2$.
Firstly, the cohomology ring $H^{*}(\mathbb{C}P^2 \# \mathbb{C}P^2)$ consists of $\langle x,y \rangle$ where $x^3=y^3=x\cup y=x^2-y^2=0$.
Secondly, the cohomology ring $H^{*}(S^2\times S^2)$ is $\langle z,t \rangle$ where $z^2=t^2=0$.
Suppose that $f^{*}\colon H^2(\mathbb{C}P^2 \# \mathbb{C}P^2)\longrightarrow H^2(S^2\times S^2)$ sends $x$ to $az+bt$ and $y$ to $cz+bt$.
Then, $0=f^{*}(x\cup y)=(ad+bc)z\cup t$, so $ad+bc=0$. Moreover, $f^{*}(x^2)=2ab\hspace{0.1cm}z\cup t$ and $f^{*}(y^2)=2cd\hspace{0.1cm}z\cup t$, but $f^{*}(x^2)=f^{*}(y^2)$, so $ab=cd$.
Furthermore, $x^2=k[\mathbb{C}P^2 \# \mathbb{C}P^2]^{*}$, so $k\hspace{0.1 cm}\text{deg }f\hspace{0.1 cm} z\cup t= 2ab \hspace{0.1 cm}z\cup t$. Hence, $2$ divides $\text{deg }f$.
Now I do not know how to proceed. I think it would be helpful to know which is the fundamental class of $\mathbb{C}P^2 \# \mathbb{C}P^2$.
Can anyone give me a hint? Thank you.
 A: There are many different continuous maps $f: S^{2} \times S^{2} \rightarrow \mathbb{CP}^{2} \#  \mathbb{CP}^{2}$, but all of them have degree zero. This follows from the equations that you wrote down.
we have $$ ad+bc=0 $$ and $$ ab = cd .$$
Without loss of generality suppose $a \neq 0$. Then we rearrange the first equation to obtain $$d =\frac{-bc}{a} $$ substituting into the second, we obtain we get $$ab =  \frac{-bc^{2}}{a},$$ hence we must have $b=0$, since otherwise we can divide both sides by $b$ to obtain $a^{2} = - c^{2}$, which contradicts the assumption $a \neq 0$. 
Hence we have that $deg(f) = 2ab = 0$ ($deg(f) = 2ab $ follows from the fact that the fundamental class of $\mathbb{CP}^{2} \#  \mathbb{CP}^{2}$ is $x^{2}=y^{2}$ so $k=1$ in your notation).
The same argument clearly works if we suppose any of $a,b,c,d$ is non-zero, hence the degree is always zero.
Here is a description of the nontrivial contiuous maps. Consider the projection $\pi_{1} : S^{2} \times S^{2} \rightarrow S^{2}$ to the first  factor and  $\pi_{2} : S^{2} \times S^{2} \rightarrow S^{2}$ to the second factor. 
Consider a projective line $L \subset \mathbb{CP}^{2}$, which is a generator of $H_{2} (\mathbb{CP}^{2}, \mathbb{Z} )$, note that this is topologically a $2$-sphere. Let $L_{1},L_{2}$ denote the projective lines inside each of the factors of $\mathbb{CP}^{2} \# \mathbb{CP}^{2}$ avoiding where the connect sum is made. These two sphere correspond to generators $x,y \in H^*(\mathbb{CP}^{2},\mathbb{Z}) $ via Poicare duality.
For $j=1,2$; given any continuous map $g: S^{2} \rightarrow L_{j} = S^{2}$, we may form a map $i \circ g \circ \pi_{i} : S^{2} \times S^{2} \rightarrow \mathbb{CP}^{2} \# \mathbb{CP}^{2}$, where $i$ is just the inclusion of either $L_{j}$ into  $ \mathbb{CP}^{2} \# \mathbb{CP}^{2}$. Clearly these maps are all degree $0$, since they factor through $S^{2}$ and $H^{4}(S^{2},\mathbb{Z})=0$. Then the above calculation shows that any continuous map must have the same induced map on cohomology as one of these maps.
