/sci/ - Science & Math

>>8681631
>>8681647
Sorry, I haven't been talking to anyone except sending my homework to a prof, not even posting here, since late December. so my communication skills have degraded quite a lot. I must also admit I assumed the reader to be familiar with these concepts, considering the fact we are discussing the Yoneda embedding here.

1) You know that isomorphism is transitive, that is, [math]X \cong Y[/math] and [math]Y \cong Z[/math] imply [math]X \cong Z[/math]? This now gives that if the stuff between the embeddings are correct, then the embeddings are isomorphic.

2) A category being cartesian closed means [math]\text{Hom}(A \times B, C) \cong \text{Hom}(B, C^A)[/math] for all objects [math]A, B, C[/math]. This is just a definition.

3) The functor [math]\text{Hom}( \cdot , X)[/math] satisfies [math]\text{Hom}(A+B, X) \cong \text{Hom}(A, X) \times \text{Hom}(B, X)[/math]. Here [math]A+B[/math] is the coproduct of [math]A, B[/math], and since the functor is contravariant, it takes coproducts to products, and vice versa.

Now, the first isomorphism (equality) is just what the embedding does. The second isomorphism is the cartesian closedness, the third is contravariance, the four is also contravariance and cartesian closedness. It could be replaced with [math]\text{Hom}(B, X^A) \times \text{Hom}(C, X^A) \cong \text{Hom}(A \times B, X) \times \text{Hom}(A \times C, X) \cong \text{Hom}((A \times B) + (A \times C), X)[/math]. Since the middle stuff is OK, the flanking objects are isomorphic, by transitivity.