Show that any map $f: S^{a + b} \to S^a \times S^b$ induces a zero map on the homology $H_{a + b}$.
I'm trying to prove this statement. My thought is that: Showing this is equivalent to showing that $f$ induces a zero map on the $H^{a+b}$ by the duality theorems in Hatcher (see Proposition 3.34 and remarks around) (is this claim correct?) Notice that composing with the projection map on $S^a$, we have $$ f: S^{a + b} \to S^{a} \times S^b \to S^a$$ which shows that the generator of $H^{a + b}(S^{a + b})$ is the pullback of the generator of $H^{a + b}(S^{a})$, which is zero. Hence the result, but I'm not sure if I can use the duality theorem in this way.
Any help or comments are welcomed!!
There are many ways to approach this problem.
One is to use the naturality of the Hurewicz homomorphism. This assumes known the standard calculation $H_nS^n\cong\mathbb{Z}$, and that $\pi_{n+k}S^n$ is torsion for $k\geq 1$. Choose, for each $n$, generators $s_n\in H_nS^n\cong\mathbb{Z}$ and for a space $X$ define the Hurewicz homorphism $h_n^X:\pi_nX\rightarrow H_nX$ by $h^X_n(\alpha)=\alpha_*s_n$, where $\alpha:S^n\rightarrow X$ represents a homotopy class in $\pi_nX$ and $\alpha_*:H_nS^n\rightarrow H_nX$ is the induced map on homology. For each space $X$ the maps $h^X_n$ are indeed group homomorphisms, as one should check.
Now observe that for a based map $f:X\rightarrow Y$ we have
$$f_*(h_n^X(\alpha))=f_*\alpha_*s_n=(f\circ\alpha)_*s_n=h^Y_n(f_*\alpha)$$
and also that $h^{S^n}_n:\pi_nS^n\rightarrow H_nS^n$ is surjective, since $h^{S^n}_n(id_{S^n})=s_n$.
Turning to the case at hand, let $f:S^{a+b}\rightarrow S^a\times S^b$ be a map, which we may assume without loss of generality to be basepoint preserving. We find that
$$f_*s_n=f_*h^{S^{a+b}}_{a+b}(id_{S^{a+b}})=h^{S^a\times S^b}_n(f_*id_{S^{a+b}})=h^{S^a\times S^b}_n(f).$$
This element must be zero for the following reason: we know that $H_n(S^a\times S^b)\cong\mathbb{Z}$ is torsion free, and also that $\pi_{a+b}(S^a\times S^b)\cong\pi_{a+b}(S^a)\oplus\pi_{a+b}(S^b)$ is torsion. Hence $f_*s_n$ lies in the image of the homomorphism $h_{a+b}^{S^a\times S^b}:\pi_{a+b}(S^a\times S^b)\rightarrow H_n(S^a\times S^b)$, which has torsion domain and free codomain so must be the trivial homomorphism. This proves the claim.
Another approach is through duality, as you describe. We will need to assume known the cup product structures on $H^*S^{a+b}$ and $H^*(S^a\times S^b)$. We know $H_nS^n\cong \mathbb{Z}$ and $H_{n-1}S^n=0$. Hence $H^nS^n\cong Hom(H_nS^n,\mathbb{Z})\cong\mathbb{Z}$ and the pairing $\langle-,-\rangle: H^nS^n\otimes H_nS^n\rightarrow \mathbb{Z}$, $\langle\alpha,s\rangle s=\alpha(s)$ is non-singular
Thus given $f:S^{a+b}\rightarrow S^a\times S^b$ we have that for $\alpha\in H^{a+b}(S^a\times S^b)$ and $s\in H_{a+b}S^{a+b}$
$$\langle\alpha,f_*s\rangle=\langle f^*\alpha,s\rangle.$$
We conlcude by non-singularity that $f_*s=0$ if and only if $f^*\alpha=0$ for all $\alpha\in H^{a+b}(S^a\times S^b)$. That is, $f_*=0$ if and only if $f^*:H^{a+b}(S^a\times S^b)\rightarrow H^{a+b}S^{a+b}$ is the zero homomorphism.
Now it is not correct that the generator of $H^{a+b}(S^a\times S^b)$ is the pullback of the generator of $H^{a+b}(S^a)$, since the latter group is trivial. However, as was pointed out in the comments, we can apply the Kunneth theorem and conclude that for generators $s_a\in H^aS^a$ and $s_b\in H^bS^b$ we have that $pr_a^*s_a=s_a\times 1$ generates $H^a(S^a\times S^b)\cong\mathbb{Z}$, that $pr_b^*s_b=1\times s_b$ generates $H^b(S^a\times S^b)\cong\mathbb{Z}$, and that $pr_a^*s_a\cup pr_b^*s_b=s_a\times s_b$ generates $H^{a+b}(S^a\times S^b)\cong\mathbb{Z}$.
It follows from all this that
$$f^*(pr_a^*s_a\cup pr_b^*s_b)=f^*(pr_a^*s_a)\cup f^*(pr_b^*s_b)=0$$
since there all cup products in $H^*S^{a+b}$ are trivial. We conclude that $f^*=0$ and get the result.