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
- $(L,\delta_A)$ is a conde of the diagram $d$ (which is trivial)
- 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:
- $C = A$
- $C \neq A$
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:
- 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 $f:B\to A$ is a cone over this diagram. So the condition for $i:L \to A$ to be a limit is that for every $f:B\to A$ there exists a unique $g:B\to L$ 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 $i:L\to A$ is a limit, since $i:L\to A$ 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.