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In a free group $F$ with basis $x_1, \ldots, x_{2k}$, how do I see that the product of commutators$$[x_1, x_2] \ldots [x_{2k - 1}, x_{2k}]$$is not equal to a product of fewer than $k$ commutators $[v_i, w_i]$ of elements $v_i$, $w_i \in F$?

Progress. If it helps, I can show that if $\Sigma_g$ denotes the closed orientable surface of genus $g$, then degree $1$ maps$$\Sigma_g \to \Sigma_h$$ exist if and only if $g \ge h$.

I know that the $2$-cell of $\Sigma_k$ is attached by the product$$[x_1, x_2] \ldots [x_{2k - 1}, x_{2k}].$$From a relation$$[x_1, x_2]\ldots[x_{2k - 1}, x_{2k}] = [v_1, w_1] \ldots [v_j, w_j]$$ in $F$, perhaps we could construct a degree $1$ map $\Sigma_j \to \Sigma_k$?

But I am stuck from here on out. Could anybody help me finish?

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    $\begingroup$ Why has this got two votes to close? It seems a good question to me! $\endgroup$
    – Derek Holt
    Nov 19 '16 at 13:21
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Denote $W_n$ to be the wedge $\bigvee_n S^1$ of $n$ circles. Define $W_{2j} \to W_{2k}$ by enumerating circles using numbers $1$ to $2j$, and sending the odd numbered circles to $v_i$, and the even ones to $w_j$. (since $v_i, w_i$ are words in $x_1, x_2, \cdots, x_{2k-1}, x_{2k}$, they define maps $S^1 \to W_{2k}$ by looking at a representative of that word in the fundamental group $\pi_1(W_{2k})$).

This map sends commutator to commutator as $[v_1, w_1] \cdots [v_j, w_j] = [x_1, x_2] \cdots [x_{2k-1}, x_{2k}]$. Hence, it extends to a map $\Sigma_j \to \Sigma_k$ which is a homeomorphism on the interior of the 2-cell. By local degree theorem, it's degree 1. That's your desired map.

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  • $\begingroup$ Welcome back :) $\endgroup$ Nov 20 '16 at 6:49
  • $\begingroup$ What's the detailed reason for being able to extend the map $\bigvee_{2j}S^1\to \bigvee_{2k}S^1$ to $M_j\to M_k$? I know at $\pi_1$ level, it does. But why on topological space level? $\endgroup$
    – Ivan So
    Aug 18 at 21:21

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