Creating a counterexample in category theory

Workout $10$ in Geroch's Mathematical Physics asks whether straight items disperse over straight amounts in approximate groups. (They carry out in the group of collections, which is what encourages the inquiry). That is, (making use of $A \times B$ to suggest the straight item and also $A+$B to be the straight amount), whether there is an isomorphism in between $A \times (B+C)$ and also $(A\times B)+(A \times C)$.

An earlier inquiry asked you to show that associativity of items holds. $(A \times B) \times C$ is isomorphic to $A \times (B \times C)$. This holds true, also when not all things have items (i.e. it holds true that if $A \times (B \times C)$ and also $(A \times B) \times C$ exist, after that they are isomorphic ...)So I take this inquiry to not call for that all things have items ...

I do not think there is the pertinent isomorphism, yet I'm not exactly sure what I've done confirms it. So my inquiry is: Is the adhering to argument an excellent way of coming close to category theory inquiries?

So I drew up a $A,B,C,A \times B,A \times,C,B+C,A\times (B+C),(A\times B)+(A\times C)$ and also reeled in all the arrowheads I recognize date the interpretations of items and also amounts (e.g. $A \times B$ warranties that there is an arrowhead from $A \times B$ to $A$ and also one to $B$ ...)Then I observed that this layout has no arrowheads that go from $A\times (B+C)$ to $(A\times B)+(A \times C)$ or the various other means. That is, also permitting make-up of arrowheads, there need not be a morphism in between both over stated things. So, I claimed, this suggests that this layout is a layout of a group where $A\times (B+C)$ is not isomorphic to $(A\times B)+(A \times C)$, given that if there are no morphisms in between them, there can not be any kind of isomorphisms.

Is this an excellent way of creating counterexamples in category theory? Exists any kind of strenuous conversation of "layouts" utilized in this manner? (I've heard it stated, yet I do not recognize where to look).

This is an inquiry concerning whether the approach I am making use of is an excellent one, not actually concerning the real fact value of the declaration concerned.

2019-05-05 02:12:21
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Answers: 5

Note that in your instance (regarding I recognize it, you have a group with just 8 things) your group does not have items and also coproducts (you have them just for some unique sets of things (as an example, you do not have item $(A\times B)\times(A\times C)$.

So if your workout had to do with groups with items and also coproducts, after that your instance is not a solution on it.

The most basic means to construct a counterexample is to examine some well-known groups. It will certainly be fairly valuable for researching category theory, due to the fact that you will certainly examine, what do abstract ideas suggest in concrete instances. In this instance this functions, you can look for instance group of vector rooms over set area.

An additional method is to make use of some method to construct new group by those you recognize (as an example, as was it was revealed by Robin Chapman).

An additional method is simply create some straightforward group - - - it is the method you have actually attempted.

2019-05-08 22:12:58

The group of of all teams is additionally a counterexample. Allowing $A,B,C:=\mathbb{Z}$, the LHS is $\mathbb{Z}\times (\mathbb{Z} * \mathbb{Z})$ and also the RHS is $(\mathbb{Z}\times \mathbb{Z})*(\mathbb{Z}\times \mathbb{Z})$, which are not isomorphic, as their Abelianizations are $\mathbb{Z}^3$, resp. $\mathbb{Z}^4$.

An additional instance is the group of Vector rooms. If $A,B,C$ are simply one dimensional vector rooms the measurement of the LHS and also the RHS do not concur (Note $\times\neq \otimes$).

2019-05-08 22:12:11

I am not a specialist on this subject, so forgive me if this is noticeable. I see what you have done, yet isn't this an instead "shallow" group? Products and also coproducts please global buildings because we can factor via them relative to various other morphisms. As an example, if there are arrowheads from B and also C each to A, after that instantly you have a new arrowhead from B+C to A which is not in your layout

Now below's where I shed the string : if you think items and also coproducts exist for all things A, B, and also C, must not a few other arrowheads exist in between A, B, and also C? Otherwise, after that you might be ALRIGHT, yet I simply do not recognize if you can think that. The interpretations of (carbon monoxide) items reference such various other arrowheads, so if there are none, after that aren't all the (carbon monoxide) items shallow?

2019-05-08 08:54:44

The twin to the declaration that $A\times(B+C)$ is isomorphic to. $(A\times B)+(A\times C)$ is that $A+(B\times C)$ is isomorphic. to $(A+B)\times(A+C)$. Currently if this declaration falls short in some. group $\mathcal{C}$ after that the initial falls short in the contrary. group $\mathcal{C}^{\mathrm{op}}$.

2019-05-08 08:49:02

What you have actually done is a reasonable start. You have actually attempted confirming the outcome and also you believe it isn't real. In this scenario you intend to seek a counter instance (and also simply one suffices).

My pointer is the group of non - commutative rings. You still have straight amount. Nonetheless the item is the free item. (For commutative rings it is the tensor item.)

2019-05-08 07:46:39