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

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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.

Can you please help me understand what am I getting wrong?
Thank you very much.

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  • $\begingroup$ 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. $\endgroup$
    – Connor Malin
    Commented Apr 2, 2020 at 14:34
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    $\begingroup$ 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. $\endgroup$
    – Connor Malin
    Commented Apr 2, 2020 at 14:40
  • $\begingroup$ 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? $\endgroup$
    – D. Hershko
    Commented Apr 3, 2020 at 5:43
  • $\begingroup$ This is just typically what is meant by multiplication in a graded abelian group. $\endgroup$
    – Connor Malin
    Commented Apr 3, 2020 at 13:13

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I think you're supposed to interpret it like this:

${\scriptsize\bigcirc}$ 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 $X$.

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  • $\begingroup$ Thank you, I'm sorry, I didn't get it, the property of maps applies for all X and Y, or just for all Y and X is fixed? and do I need to prove for every X? or just the fixed X? $\endgroup$
    – D. Hershko
    Commented Apr 2, 2020 at 16:11
  • $\begingroup$ 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 $\forall X\ \big({\scriptsize\bigcirc}_X = 0\big)$. $\endgroup$
    – William
    Commented Apr 2, 2020 at 16:21

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