# Help in understanding a question about bilinear multiplication on homology groups

I am having a trouble understanding the following question in algebraic topology:

My problem is that if $$f\colon X\to Y$$ then $$f_*(\alpha)$$ and $$f_*(\beta)$$ are in $$H_*(Y)$$ and not in $$H_*(X)$$, which is where the multiplication is defined.

Thank you very much.

• The problem is a little unclear. It is asking for a multiplication defined on each space such that it has this property with respect to maps. Commented Apr 2, 2020 at 14:34
• As a hint: try to come up with a space that has it's nth homology map onto your class in $H_n(X)$ and its mth homology map onto your other class in $H_m(X)$, but the $n+m$th homology of the space is trivial. Commented Apr 2, 2020 at 14:40
• thanks for the comment, did you mean that the multiplication of two classes, one in $H_n(X)$ and other in $H_m(X)$ must be in $H_{n+m}(X)$? why it must be the case? Commented Apr 3, 2020 at 5:43
• This is just typically what is meant by multiplication in a graded abelian group. Commented Apr 3, 2020 at 13:13

$${\scriptsize\bigcirc}$$ is a bilinear operation defined on the homology of every topological space. For a given $$X$$ denote the operation by $${\scriptsize\bigcirc}_X \colon H_*(X)\otimes H_*(X) \to H_*(X)$$.
Now suppose that for every pair of spaces $$X, Y$$ and every continuous function $$f\colon X \to Y$$ we have the formula $$f_*(\alpha{\scriptsize\bigcirc}_X \beta) = f_*(\alpha) {\scriptsize\bigcirc}_Y f_*(\beta)$$.
Then the problem is to prove that $${\scriptsize\bigcirc}_X$$ is trivial for all $$X$$.
• Assuming I'm interpreting it correctly, the appropriate property is "$\forall X\ \forall Y\ \forall f\colon X\to Y\ \big(f_*(\alpha{\scriptsize\bigcirc}_X \beta) = f_*(\alpha) {\scriptsize\bigcirc}_Y f_*(\beta)\big)$ and the thing you want to prove is $\forall X\ \big({\scriptsize\bigcirc}_X = 0\big)$. Commented Apr 2, 2020 at 16:21