Cfrgtkky

I read from these slides:

a limit of a diagram containing just one object A and no
morphism is any object L that is isomorphic to A (the
isomorphism is part of the limit);

and was unsure why we needed the isomorphism between $A$ and $L$. Isn't $L$ in this scenario a limit even if A and L are not isomorphic because there is no other cone (so that condition is vacuously satisfied?).

i.e. if L points to diagram A, isn't L a limit of that diagram?

How does the isomorphism play a role in showing that $L$ is a limit of the diagram $A$?

My thoughts:

The way I understand the question is as follows. The claim is that a limit of a diagram with a single object $A$ is any $(L,delta_A)$ such that $L cong A$. So if $L cong A$ then it is a limit. To show $L$ is a limit we need to show that

1. 
   $(L,delta_A)$ is a conde of the diagram $d$ (which is trivial)


2. for any other cone $(C,gamma_A)$ that there exists a unique factorization given by $gamma_A = h; delta_A$.

In particular I believe there are two cases to check:

1. $C = A$


2. $C neq A$

I think for case 1 part of the proof goes as follow: Consider any cone $(C,gamma_A)$. Then for $(L,delta_A)$ to be a limit we need to check $gamma_A = h; delta_A = delta circ h$ (assume we already know they are cones of $d$). Since we know $L cong A$ we know there exists $f: L to A$ and $g: A to L$ such that $f;g = 1_L$ and $g;f = 1_A$. Therefore to satisfy $gamma_A = h; delta_A$ choose the $h$ that makes $1_A= h; delta_A$ true. Here is where I get confused:

1. How do I know $gamma_A = 1_A$? Is this where the assumption in the paragraph I quoted goes into play? i.e. "the isomorphism is part of the limit". How do we know $delta_A$ does have that property? $delta_A$ is a morphism from $L to A$. If this was in the category Set it could be that $delta_A$ maps everything from $L$ to a single element of $A$, which makes it non-invertible, so I can't see how the bijection could take place.

the second case is even less clear to me how we show such a factorization exists and what is the role of saying $L$ is isomorphic to $A$ is.

There are lots of cones! For any object $B$, every morphism $f:Bto A$ is a cone over this diagram. So the condition for $i:L to A$ to be a limit is that for every $f:Bto A$ there exists a unique $g:Bto L$ such that $ig=f$. Taking $B=A$ and $f=1_A$ gives a morphism $g$ such that $ig=1_A$. We then have $igi=i=i1_L$ which implies $gi=1_L$ using the uniqueness condition with $f=i$. So, $i$ is an isomorphism with inverse $g$.

I would also remark that even if there are no other cones, that still doesn't immediately imply that $i:Lto A$ is a limit, since $i:Lto A$ is still a cone and so the definition is not vacuous. The limit condition is required to hold for all cones, not just those that are different from the limiting cone.