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>> No.11596320 [View]
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11596320

>>11595741
I'm personally used to seeing [math]O(3, 1)[/math] as Lorentz and [math]SO^+(3, 1)[/math] as proper Lorentz.
>>11596258
Right, so if we call the two categories [math]A[/math] and [math]B[/math], there *is* an induced map [math]F^*: \mathscr{F}_A \rightarrow \mathscr{F}_B[/math] between the free groups. This is because isomorphic objects have isomorphic images under functors. Also, there's a projection map [math]\pi: \mathscr{F}_B \rightarrow \mathscr{F}_B / \mathscr{R}_B[/math].
The composition gives a map [math]\pi \circ F^* : \mathscr{F}_A \rightarrow \mathscr{F}_B / \mathscr{R}_B[/math].
To check if it's defined on [math]\mathscr{F}_A / \mathscr{R}_A[/math], it's necessary and sufficient that [math]ker \pi \circ F^* \supseteq \mathscr{R}_A[/math]. For that, it's sufficient to check that the generators are in the kernel. And then that follows because of the exactness.
I think it works.

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