I am reading the following paper of Miller: http://researchers.ms.unimelb.edu.au/

He says that if $G= F_{1} \times F_{2}$ is a direct product of two free groups and $H$ is a subgroup of $G$, then it can be assumed that the projection maps $$\text{p}_{i} \colon H \rightarrow F_{i}$$ are surjective.

I do not understand why is this trivial. If $\{f_{1},\cdots,f_{n}\}$ is a basis of the free group $F_{1}$, why should we have elements of the form $(f_{i},h_{i})$ in $H$ for all $i\in \{1,\cdots,n\}$? He simply says that this follows because subgroups of free groups are free.

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    $\begingroup$ Let $H_i\subset F_i$ be the image of $H$ under $p_i$. Then $H_i$ is free and $H\subset H_1\times H_2$ $\endgroup$ – Max Jan 3 at 17:37
  • $\begingroup$ Omg it was trivial and I have been ages thinking about it! Thank you very much @Max :) $\endgroup$ – Karen Jan 3 at 17:44
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    $\begingroup$ Don't beat yourself up about it, @Karen; it took over three hundred pages of dense symbol shuffling to get to a proof that, in fact, yes, $1+1=2$ in Russell & Whitehead's "Principia Mathematica". $\endgroup$ – Shaun Jan 3 at 17:51

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