Algebra 3 | Identities and Inverses

[@sinx2247]

10:55 Does the fact that surjective functions have a right inverse depend on the axiom of choice? To construct such a right inverse, we need to be able to choose an element from the preimage of f for every element in its codomain

[@brightsideofmaths]

Yes, it's equivalent to the axiom of choice.

[@thes7274473]

Wouldn't it be possible to have multiple one-sided neutral elements so long as they're all of the same side? Say we have a set X closed under an operation ∘. If e₁ ∘ x = x and e₂ ∘ x = x for all x ∈ X. e₁ ∘ e₂ = e₂ and e₂ ∘ e₁ = e₁ does not necessitate that e₁ = e₂ because e₁ ∘ e₂ ≠ e₂ ∘ e₁

[@brightsideofmaths]

You can just create an example for that :)

[@gavasiarobinssson5108]

Difficult to imagine two different neutral elements. Are there examples?

[@brightsideofmaths]

Thinking about your claim, you could just consider a set with two elements {a,b} and define: a ∘ a = a, a ∘ b = b, b ∘ a = a, and b ∘ b = b

[@gavasiarobinssson5108]

@@brightsideofmaths i asked chatgpt and he suggested max(a,b) with a and b belonging to Z+ and including minus infinity. Then both zero and minus infinity are left side neutral elements. But I dont think so

[@Happy_Abe]

@@gavasiarobinssson5108I don’t think that’s right since if 0=e1 and negative infinity=e2 Then max(e1,e2)=e1 so e1 is not a left identity.

[@gavasiarobinssson5108]

Great!

[@Happy_Abe]

Is there a result that states if an element has a left inverse x and a right inverse y then x=y? That’s true in groups, but is it true in semigroups since the existence of the left and right inverses isn’t guaranteed? Similarly, are left inverses unique? If xa=e=ya, does that imply that x=y? Do we get any kind of inverse uniqueness if we’re not in a group?

[@JulianEpsilon]

>Is there a result that states if an element has a left inverse x and a right inverse y then x=y? let a_l be the left inverse of a and a_r be the right inverse. a_l ∘ a = e e = a ∘ a_r a_l ∘ e = a_l ∘ a ∘ a_r a_l ∘ e = ( a_l ∘ a ) ∘ a_r a_l ∘ e = ( e ) ∘ a_r a_l = a_r > Similarly, are left inverses unique? in the funcion exaample we see 0.5sqrt(x) +0.5 is a right inverse of 4(x-1/2)^2 which means 4(x-1/2)^2 is a left inverse of 0.5sqrt(x) +0.5 but it doesn't matter what the left inverse of (0.5sqrt(x) +0.5) does on the domain of [0, 0.5) so h(x) = { whatever(x) , x ∈ [0, 0.5) { 4(x-1/2)^2, x ∈ [0.5, 1] is a family of left inverses of (0.5sqrt(x) +0.5)

[@Happy_Abe]

@@JulianEpsilonthanks for your answers! Great answer for the first part. It seems to follow from the associative structure of the binary operation and if the operation wouldn’t be associative this may fail. For the second question I’m not sure I understand what you’re trying to say. Why wouldn’t the function matter for the domain [0, .5), wouldn’t we want the inverse to work everywhere on the original domain of [0,1] Thanks again!