Show a map in cohomology is trivial Let $X=S^{1}\vee S^{2}\vee S^{3}$ and $Y=S^{1}\times S^{2}$. I already compute
$$ H^{k}(X)=\begin{cases}
\mathbb{Z}, \ \ \text{if} \ k=0,1,2,3 \\
0, \ \ \text{if} \ k>3 
\end{cases} \ \ \ \ \ \text{and} \ \ \ \ \
H^{k}(Y)=\begin{cases}
\mathbb{Z}, \ \ \text{if} \ k=0,1,2,3 \\
0, \ \ \text{if} \ k>3 
\end{cases}$$
How can I prove that for any continuous map $f:X\to Y$ the induced map in cohomology $f^{*}:H^{3}(Y)\to H^{3}(X)$ is trivial.
My idea is use the cup product as following:
$\require{AMScd}$
\begin{CD}
H^{1}(Y)\otimes H^{2}(Y) @>{\smile}>> H^{3}(Y)\\
@VVV @VVV\\
H^{1}(X)\otimes H^{2}(X) @>{\smile}>> H^{3}(X)
\end{CD}
How can I continue? I will apreciate any hint to solve this.
 A: Let $\alpha$ and $\beta$ be generators of $H^1(Y)$ and $H^2(Y)$ respectively. Then $\alpha \smile \beta$ is a generator of $H^3(Y)$. (This follows easily from the Kunneth formula, since if $p_1 : Y \to S^1$ and $p_2 : Y \to S^2$ are the natural projections and $\widetilde \alpha$ and $\widetilde \beta$ are generators of $H^1(S^1)$ and $H^2(S^2)$ respectively, then Kunneth implies that $p_1^\star(\widetilde \alpha)=p_1^\star(\widetilde \alpha)\smile p_2^\star(1)$ , $p_2^\star(\widetilde \beta)= p_1^\star(1) \smile p_2^\star(\widetilde \beta)$ and $p_1^\star(\widetilde \alpha) \smile p_2^\star(\widetilde \beta)$ are generators of $H^1(Y)$, $H^2(Y)$ and $H^3(Y)$ respectively.)
So let's consider the pullback $f^\star(\alpha \smile \beta)$ of the generator $\alpha \smile \beta \in H^3(Y)$.  This is equal to $f^\star(\alpha) \smile f^\star(\beta)$. As $X$ is a wedge sum of $S^1$, $S^2$ and $S^3$, we have a ring isomorphism $\widetilde H^\star (X)\cong\widetilde H^\star(S^1)\oplus \widetilde H^\star(S^2)\oplus\widetilde H^\star(S^3),$ (where the tildes indicate reduced cohomology, i.e. ignoring the generator of zeroth cohomology). Via this ring isomorphism, $f^\star(\alpha)$ maps to an element of the form $ (\alpha', 0, 0)$ (where $\alpha' \in \widetilde H^1(S^1)$), while $f^\star(\beta)$ maps to an element of the form $(0, \beta', 0)$ (where $\beta'\in \widetilde H^2(S^2)$). Hence $f^\star(\alpha) \smile f^\star(\beta)$ maps to $(\alpha' \smile 0, 0 \smile \beta', 0 \smile 0) = (0,0,0)$ via this ring isomorphism, which shows that $f^\star(\alpha \smile \beta)$ is zero, as required.
