If $\mathscr A$ has all products and equalizers, then it has all limits

Question

This question is about part (a) of this proposition:

Here is a plan of the proof.

Here's what the picture looks like, from what I understand:

But I don't understand what the condition $s\circ p=t\circ p$ means. I guess this means componentwise equality: $s_u\circ p_u=t_u\circ p_u $ for all $u$, but what is $p_u$? $L$ is not a product, and so $p_u$ cannot be a projection.

I'm just trying to prove first that $Du\circ p_I=p_J$ for all $u: I\to J$, and it seems I need to use $s\circ p=t\circ p$ to that end, which is unclear how to do.

Answer 1

$s \circ p = t \circ p$ are both arrows into a product, the equality here is equivalent to them being equal after each projection, i.e., $\pi_u \circ s \circ p = \pi_u \circ t \circ p$ for all $u : J \to K$ in $\mathbf I$.

Now, $\pi_u \circ s$ is the $u$-component of $s$, hence is equal to $D(u) \circ \text{pr}_J$. Similarly, $\pi_u \circ t = \text{pr}_K$. Combining this with the previous equality, we find $D(u) \circ \text{pr}_J \circ p = \text{pr}_K \circ p$. Now, by definition $p_I = \text{pr}_I \circ p$ for any $I \in \mathbf I$ (i.e. $p_I$ is the $I$-component of $p$), we find $D(u) \circ p_J = p_K$.

Hence $(L, (p_I : L \to D(I))_{I \in \mathbf I})$ is a cone. Hopefully this clears up the confusion.

Answer 2

This is the second part of the proof (the first part is Rushy's answer).

We need to show that if $(L', f_I:L'\to D(I))_{I\in\mathbf{I}}$ is another cone, then there exists a unique arrow $\alpha: L'\to L$ such that $p_J\circ\alpha=f_J$ for all $J\in\mathbf I$.

So let $(L', f_I:L'\to D(I))_{I\in\mathbf{I}}$ be a cone. By the universal property of the product, there exists a unique arrow $f: L'\to \prod_{I\in\mathbf I} D(I)$ such that $pr_J\circ f=f_I$. (In other words, the $I$th component of $f$ is $f_I$.)

Note that $s\circ f=t\circ f$. Indeed,

$$s\circ f=t\circ f \\ \iff \pi_u\circ s\circ f=\pi_u\circ t \circ f \text{ (see Rushy's comment under the answer)}\\ \iff s_u\circ f=t_u\circ f \text{ (definition of the }u\text{th component of $s$ and $t$)} \\ \iff Du\circ pr_J\circ f =pr_K\circ f \text{ (definitions of $t_u$ and $s_u$)} \\ \iff Du\circ f_J=f_K \text{ (by the above)}$$ The equality in the last row holds because $(L', f_I:L'\to D(I))_{I\in\mathbf{I}}$ is a cone. Thus $s\circ f=t\circ f$.

Since $(L,p:L\to \prod_{I\in\mathbf I} D(I))$ is an equalizer of $s$ and $t$, there is a unique $\alpha: L'\to L$ such that $p\circ\alpha=f$. Again, by the answer in the comments, this is equivalent to $pr_J \circ p\circ \alpha=pr_J\circ f$. Since $pr_J\circ \alpha=p_J$ and $pr_J\circ f=f_J$, we can rewrite that as $p_J\circ \alpha=f_J$. That's what we needed to prove.