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H . ✠ . ❄ ❄❄✙ ... H z ✲ K such that ✲ Q .. B Dually an epimap is regular iff it is the coequalizer of the relation it induces on its domain. Proposition 3. If k1 is a monomorphism and k2 k1 is a regular monomap, then k2 is a regular monomap, and dually. 3 Regular epimorphisms and monomorphisms Proof. Consider 60 q3 = (k2 k1 j1 )E ∗ (k2 k1 j2 ) k2 k1 = (j1 q3 )E(j2 q3 ) q2 = (k2 i1 )E ∗ (k2 i2 ) .. . z.... K2 . ✒ X k2 ❅ ❅ x ❅ ❅ ❘ ❅ ❄ K1 i1 k1 j1 i2 ❄❄ K1 K1 ✲ A k1 k1 q2 ✲A j2 ❄❄ A q3 ❄ Q2 v ❄ ✲ Q3 We must show k2 = (i1 q2 )E(i2 q2 ).

Is a candidate for k2 k1 . z[xk1 = zk2 k1 ]. But k1 is a monomorphism, so x = zk2 . Finally, z satisfying the last equation is unique, since k2 is a monomorphism. Chapter II Algebraic theories 1. The category of algebraic theories Before discussing the category of algebraic theories, we briefly consider the category ({S0 }, C1 ), of which the category of algebraic theories will be a subcategory. Recall that the maps in the category ({S0 }, C1 ) may be identified as commutative triangles S0 A A ✠ f ❅ ❅ B ❅ ❅ ❘ ❅ ✲ B of functors where S0 is (any fixed small version of) the category of finite sets and A, B are any small categories.

If k1 is a monomorphism and k2 k1 is a regular monomap, then k2 is a regular monomap, and dually. 3 Regular epimorphisms and monomorphisms Proof. Consider 60 q3 = (k2 k1 j1 )E ∗ (k2 k1 j2 ) k2 k1 = (j1 q3 )E(j2 q3 ) q2 = (k2 i1 )E ∗ (k2 i2 ) .. . z.... K2 . ✒ X k2 ❅ ❅ x ❅ ❅ ❘ ❅ ❄ K1 i1 k1 j1 i2 ❄❄ K1 K1 ✲ A k1 k1 q2 ✲A j2 ❄❄ A q3 ❄ Q2 v ❄ ✲ Q3 We must show k2 = (i1 q2 )E(i2 q2 ). Now k2 ‘equalizes’ i1 q2 , i2 q2 by definition of q2 . e. is a ‘candidate’ for q2 . e. is a candidate for k2 k1 . z[xk1 = zk2 k1 ].

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A Quick Review Of Commutative Algebra [Short Lecture] by Ghorpade


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