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I think this argument works in the case that the inner product is definite (which is the case you are considering anyways). In this case, the Pin group is generated by the set of all normalized non-zero vectors. Therefore, by the universal property of Cl, a rep of Pin gives a unique rep of Cl and restricting this rep gives you back the rep of Pin you started with since they agree on unit vectors. Thus every rep of Pin is the restriction of a rep of Cl. If you know where all the $e_i$'s get mapped, you then get a unique rep of Cl which gives a unique rep of Pin.

If the inner product is not definite you will still get a rep of Cl but I can't see how you know that it will agree with the original rep of Pin.

EDIT: I have assumed that the map we get from the set of all unit vectors (by restricting the map from Pin), is the restriction of a linear map from $R^n$. This doesn't seem to be obviously the case, but is equivalent to the statement that every rep of Pin is the restriction of a rep of Cl.

Post Undeleted by Eric O. Korman
I think I may be missing something but: if this argument works in the case that the inner product is definite (which is the case you have a rep of are considering anyways). In this case, the Pin then you get a rep group is generated by the set of all normalized non-zero vectors. Therefore, by the third groupuniversal property of Cl, which a rep of Pin gives a unique rep of Cl by and restricting this rep gives you back the universal property rep of Cl (as Michael mentioned in his comment)Pin you started with since they agree on unit vectors. Thus every rep of Pin is the restriction of a unique rep of Cl. If you have two reps of Pin that agree on know where all the $e_i$, e_i\$'s get mapped, you then they lift to get a unique rep of Cl which gives a unique rep of Pin.