Multiplication in homology groups which satisfies a condition is trivial Let $H_*(X)$ denote the union of all homology groups of $X$.
Let $\circ$ be a bilinear multiplication on all $H_*(X)$  (defined for all $X$).
I.e. for all $X$ ,  $\circ _X: H_*(X)\times H_*(X)\to H_*(X)$ such that if $[a],[b]\in H_n(X)$ and $[c]\in H_m(X)$ than $([a]+[b])\circ _X [c] = [a]\circ _X [c] + [b]\circ _X [c]$, and the same thing for the second coordinate.    
we define $\mathbf {condition}$ for $\circ$: For every $X,Y$ , $[\alpha],[\beta]\in H_*(X)$, every $f:X\to Y$ satisfies $f_*([\alpha]\circ _X[\beta])=f_*([\alpha])\circ _Y f_*([\beta])$ 
Prove that every bilinear multiplication which satisfies the condtion above, must be trivial, i.e. $[\alpha]\circ _X[\beta]=0$ for every $X$, and $[\alpha],[\beta]\in H_*(X)$.  
There is a hint to look at the inclusions $in1,in2:X\to X\lor X$ and the projections $pr1,pr2:X\lor X\to X$.  
I don't know what to do to solve, I would be very happy for a hint or a solution.
Thank you very much.
 A: HINT:
Suppose that you have a multiplication which satisfies your condition $(*)$:

For every $X,Y, [\alpha],[\beta]\in H_*(X)$ every $f:X \rightarrow Y$
  satisfies $f_*([\alpha]\circ _X[\beta])=f_*([\alpha])\circ _Y f_*([\beta])$.

Firstly, let $Y = X \vee X$, and $f_{1,2}:X \rightarrow X \vee X$ be the inclusions into the first and second factor, respectively. Start by writing down exactly what maps these inclusions induce on the level of homology. Then do the same with the projection maps (letting $X = X \vee X$ and $Y = X$).
Now you should use the fact that you have condition $(*)$. Before reading further, I would advise that you think about how to do that; you're told to assert a condition, you then want to prove a property. What exactly does the condition tell you, and how do you exploit it?
As hinted in the comments, you can start with proving that $(in_1)_* x\circ_{X\vee X}(in_2)_*y = 0$. Now notice what this actually says (hint: any $(x,y) \in X \vee X$ can be written as $((in_1)_* x,(in_2)_*y )$).
Now you can continue by looking at the composition $(pr_1)(in_1): X \rightarrow X$.
Applying that composition to an appropriate thing (keeping in mind that what you're trying to prove is that the multiple of two arbitrary elements is trivial) should give you what you want.
