Flaw in a proof of uniqueness of products in a category?

Question

I previously thought I knew a proof of the unique "upto unique isomorphism" of the product in a category. I was recently presented with a more complicated proof in a textbook. I am therefore left to wonder whether my "simpler" proof is somehow subtly wrong.

A product of $A$ and $B$ in a category $C$ is a 3 tuple $(P \in C, \pi^p_a \in Hom(P, A), \pi^p_b \in Hom(P, B))$ such that for any other 3-tuple ($Q \in C, \pi^q_a \in Hom(Q, A), \pi^q_b \in Hom(Q, B))$, we have a unique morphism $q2p: Hom(Q, P)$ such that $\pi^q_a = \pi^p_a \circ q2p$, and $\pi^q_b = \pi^p_b \circ q2p$.

Now, we wish to show that given two products $K, L$ of $A$ and $B$, that there is a unique isomorphism between $K$ and $L$. That is, we have two maps $k2l \in Hom(K, L)$ and $l2k \in Hom(l, k)$ such that $k2l \circ l2k = id_l$ and $l2k \circ k2l = id_k$.

I thought the proof of uniqueness of the product goes like this:

1. Assume we have two candidates for the product of $A$ and $B$, namely, $(A \times B, pr_1, pr_2)$, and $(A \otimes B, pr'_1, pr'_2)$.
2. By the universal property of $A \times B$, we have a unique map $k \in Hom(A \otimes B, A \times B)$ such that $pr'_1 = pr_1 \circ k$, $pr'_2 = pr_2 \circ k$.
3. Similarly, by the universal property of $A \otimes B$, we have a unique map $l \in Hom(A \times B, A \otimes B)$ such that $pr_1 = pr'_1 \circ l$, $pr_2 = pr'_2 \circ l$
4. These together give us a map $k \circ l \in Hom(A \times B, A \times B)$ which behaves like the identity element: $pr_1 \circ id = pr_1 = pr_2 \circ (k \circ l)$, and similarly $pr_2 \circ id = pr_2 = pr_2 \circ (k \circ l)$
5. By the uniqueness of identity, we have that $id_{A \times B} = k \circ l$.

Rather, the trusted proof of product that I have seen changes at the 5th step. It proceeds as:

5. We now apply the universal property a third time with $A \times B$ on $A \times B$, telling us that there exists a unique map $h \in Hom(A \times B, A \times B)$ such that $pr_1 = pr_1 \circ h$, $pr_2 = pr_2 \circ h$.
6. We have two such candidates for such an $h$: $id_{A \times B}$ and $k \circ l$. But since $h$ is unique, we have that $id_{A \times B} = k \circ l$.

I am confused as to why we cannot conclude the proof the way I do. As I see it, since all the maps $k, l, id_{A \times B}$ are unique, they must coincide?

I suppose the flaw with my argument is that I only know that $k \circ l$ agrees with $id_{A \times B}$ at $pr_1$ and $pr_2$. There maybe other morphisms in $Hom(A \times B, A \times B)$ where $id_{A \times B}$ and $k \circ l$ may not agree. Is my identification of the error correct?

Answer 1

To check that $k\circ l$ "behaves like the identity element", one had to show that $k\circ l$ is a neutral element w.r.t. to every morphism coming from or to $A\times B$.

In your proof, you showed that $k\circ l$ is a neutral element from the right w.r.t. $pr_1,pr_2$, which isn't enough for your claim. However, as pointed out by your trusted proof, there is only one such morphism behaving as a right identity for $pr_1,pr_2$, so $k\circ l = id_{A\times B}$.