Constructing finite limit from terminal object and pullbacks

Question

I'm trying to understand, why in Proposition 3.1 in Finite Limit in ncatlab (3) implies (1). My intuition tells me to use induction to show that the limit for any diagram having $n$ vertices (objects) (with $n \geq 0$) exists:

Base case: The limit over the empty diagram ($n = 0$) is given by the terminal object.

Now suppose we already now that the limit of a diagram with $n$ vertices exists. We construct the limit of a diagram $D$ with $n + 1$ vertices as follows: We take the limit $L'$ of a sub-diagram $D'$ of $D$ with $n$ vertices. We denote the remaining vertex by $A_{n+1}$ and the terminal object by $T$. Moreover we denote the unique morphism from an object $C$ to $T$ by $t_C$. We now take the pullback $L = L' \times_T A_{n+1}$ of $L' \rightarrow T \leftarrow A_{n+1}$: $\require{AMScd}$ \begin{CD} L' \times_T A_{n+1} @>{p_2}>> A_{n+1} \newline @V{p_1}VV @VV{t_A}V \newline L' @>{t_{L'}}>> T \end{CD}

We claim that $L$ is the limit of $D$.

We have that for any morphism $g: L \to A_i$ (with $1 \leq i \leq n$), $g = g' \circ p_1$ for some $g' : L' \to A_i$, since $L'$ is a limit of $D$ (thus uniquely determining $p_1$). Also, since $T$ is the terminal object, $t_{L'} = t_{A_i} \circ g'$. Therefore $t_{L'} \circ p_1 = t_{A_i} \circ g$. Hence, if there are no morphisms between any of the $A_i$ and $A_{n+1}$, then it follows from the properties of a pullback that $L$ is indeed a limit.

Now my problem comes in: If there is some morphism $h : A_i \to A_{n+1}$, we need $p_2 = h \circ g$. How can we know that this is true? Or can we not know that and I should use a different approach?

Answer 1

Let’s look at the situation in $\mathbf{Set}$ for a moment. Given a (small) index category $I$ and a diagram $$ D \colon I \to \mathbf{Set} \,, $$ its limit $L$ can be constructed as $$ L = \Bigl\{ (x_i)_i ∈ ∏_{i ∈ I} D(i) \;\Big|\; \text{$D(u)(x_i) = x_j$ for every morphism $u \colon i \to j$ in $I$} \Bigr\} \,. $$

Let us now remove an object $i^*$ from the category $I$. That is, let $J$ be the full subcategory of $I$ that is only missing the single object $i^*$. The diagram $D$ restricts to a diagram $$ D' \colon J \to \mathbf{Set} \,, $$ which admits a limit $L'$. We have for $L'$ the same kind of explicit description as for $L$.

We find from these explicit description of $L$ and $L'$ that $$ L ≅ \left\{ (x, (y_j)_j) ∈ D(i^*) × L' \,\middle|\, \begin{array}{l} \text{$D(u)(x) = y_j$ for every object $j ∈ J$ and morphism $u \colon i^* \to j$ in $I$,} \\ \text{$D(v)(y_k) = x$ for every object $k ∈ J$ and morphism $v \colon k \to i^*$ in $I$} \end{array} \, \right\} \,. \tag{$\ast$} $$ We can therefore describe $L$ as a pullback of the following form: $$ \begin{CD} L @> \mathrm{pr}_2 >> L' \\ @V \mathrm{pr}_1 VV @VVV \\ D(i^*) @>>> \Biggl( \prod_{\text{$u \colon i^* \to j$ in $I$}} D(j) \Biggr) × \Biggl( \prod_{\text{$v \colon k \to i^*$ in $I$}} D(i^*) \Biggr) \end{CD} $$

In your approach, you chose the lower right object in this diagram as the terminal object instead. This amounts to ignoring all the conditions in $(\ast)$. You thus end up with the product $D(i^*) × L'$ instead of the limit $L$. More generally, your recursive construction results in the product $∏_{i ∈ I} D(i)$ instead of the desired limit $L$.

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I’ve also tried to find a direct way from (3) to (1), but have so far been unsucessful. (All my ideas require at least the existence of binary products at some point or the other.)