every map $S^4\to S^2\times S^2$ has degree $0$ I am working on Algebraic Topology in the past Quals.
"Prove that every map $S^4\to S^2\times S^2$ has degree $0$,
      i.e. the induced homomorphism in $H_4$ is $0$."
Here is how I approach: the first observation is that we can use Kunneth formula to compute $H_4(S^2\times S^2)=H_4(S^4)=\mathbb{Z}$. Since $S^4$ is simply connected, a map $f:S^4\to S^2\times S^2$ is lifted to a map to the universal cover $g:S^4\to S^2\times S^2$, which basically means nothing here. (this approach can be used to solve a similar problem: "any map $S^2\to S^1\times S^1$
is nullhomotopic" because the lift gives us more information.)
I'm stuck here. How do I solve this?
 A: Here's a solution. 
Translating degree into cohomology
We can translate degree into cohomological terms using the universal coefficient theorem.
We have natural short exact sequences 
$$\newcommand\Ext{\operatorname{Ext}}\newcommand\Hom{\operatorname{Hom}}
0\to \Ext^1_{\Bbb{Z}}(H_{i-1}(X),\Bbb{Z})
\to H^i(X) \to \Hom(H_i(X),\Bbb{Z})\to 0
$$
which are split, though not naturally.
If $H_i$ is always free, or at least $H_{n-1}$ is free, then the $\Ext$ terms vanish, and we have a natural 
isomorphism 
$$H^n(X)\xrightarrow{\sim} \Hom(H_n(X),\Bbb{Z}).$$
Naturality means that for $f:X\to Y$, 
$f^*:H^n(X)\to H^n(Y)$ corresponds under the isomorphism to 
$\Hom(f_*,\Bbb{Z}) : \Hom(H_n(Y),\Bbb{Z})\to \Hom(H_n(X),\Bbb{Z})$.
Note that $\Hom$ is contravariant in its first argument, so the direction makes sense.
Thus if $X$ and $Y$ have chosen fundamental classes $[X]$ and $[Y]$, so that 
$H_n(X)\cong \Bbb{Z}\{[X]\}$ and $H_n(Y)\cong \Bbb{Z}\{[Y]\}$, we know that 
$f_*[X]=k[Y]$ for some integer $k$, which is the degree of $f$. ($\Bbb{Z}\{a,b,c,\ldots\}$ is notation for the free $\Bbb{Z}$-module on the set of generators.)
Then $H^n(X)\cong \Hom(H_n(X),\Bbb{Z})$ is also isomorphic to $\Bbb{Z}$, and is  generated by the map $\alpha_X:[X]\mapsto 1$. Similarly for $Y$.
We know that $f^*\alpha_Y = k'\alpha_X$ for some integer $k'$, which we might call the cohomological degree of $f$.
In fact,
since $f^*\alpha_Y = \alpha_Y\circ f_*$, we know 
$$f^*\alpha_Y([X]) = \alpha_Y(f_*[X])=\alpha_Y(k[Y])=k\alpha_Y([Y])=k.$$
Therefore $k'=k$, so cohomological degree is the same as the usual degree, and we can just call both the degree and measure it using either homology or cohomology.
The cohomological one is more useful in our case, since we have additional structure on cohomology, namely the cup product.
Applying the cohomological version of degree to solve the problem
For cohomology, when all the Tor factors vanish, the Kunneth theorem gives an isomorphism of graded rings, not just groups. See the answers here and probably check Hatcher for a reference, though I haven't looked recently, but I think it's in there somewhere.
Now we know $H^*(S^2) \cong \Bbb{Z}[x_2]/(x_2^2)$, where $|x_2|=2$, so 
$H^*(S^2\times S^2)=\Bbb{Z}[x_2,y_2]/(x_2^2,y_2^2)$, and 
$H^*(S^4)=\Bbb{Z}[x_4]/(x_4^2)$, where $|x_4|=4$. 
Then the induced map $f^*:H^*(S^2\times S^2)\to H^*(S^4)$ is determined by where it sends $x_2$ and $y_2$, and it has no choice but to send these to $0$, since $H^2(S^4)=0$. Thus 
$f^*(x_2y_2) = f^*(x_2)f^*(y_2)=0$.
Thus the degree of $f$ is $0$.
