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Is $\mathbf{Grp}$ a concrete category? I thought it is, but then the group of symmetries of a square and the quaternion group are both of the order 8, and they are not isomorphic as groups. But sets of the same cardinality are isomorphic as sets, so a standard forgetful functor $(G, \cdot) \mapsto G$ is not faithful! Am I missing something here?

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up vote 20 down vote accepted

Yes, you have the wrong definition of a faithful functor. A functor F is faithful precisely when F sends morphisms injectively, but nowhere is there required any kind of injectivity on objects.

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Oh, thank you! :D – Alexei Averchenko Dec 12 '10 at 12:10

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