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Let's take the universal property of tensor products for example, according to Wikipedia:

Uniqueness of the tensor product means that for any other bilinear map $φ′: V × W → V ⊗'W$ with the above property there is an isomorphism $k : V ⊗ W → V ⊗'W$ such that $φ′ = k ∘ φ$ holds.

What he means by "above property" is that

...there is a bilinear map (i.e., linear in each variable v and w) $φ: V × W → V ⊗ W$ such that given any other vector space(or module) $Z$ together with a bilinear map $h : V × W → Z$, there is a unique linear map $\tilde{h}: V ⊗ W → Z$ satisfying $h = \tilde{h}∘ φ$.

My question is that, can we simply remove "unique" from here in order to use the universal property? It seems to me that once we construct $\tilde{h}$, it is always unique. To be specific, if the following statement is also true:

If for any other bilinear map $φ′: V × W → V ⊗'W$ with the property that given any other vector space(or module) $Z$ together with a bilinear map $h : V × W → Z$, there is a linear map $\tilde{h}: V ⊗' W → Z$ satisfying $h = \tilde{h}∘ φ'$, then there is an isomorphism $k : V ⊗ W → V ⊗'W$ such that $φ′ = k ∘ φ$ holds.

But my question is not just limited to the case for tensor products. I wonder if there is a category where the proofs of both existence and uniqueness are necessary.

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Uniqueness is absolutely necessary. For instance, let $U$ be any vector space, and define $V\otimes' W=(V\otimes W)\oplus U$. There is a bilinear map $\varphi':V\times W\to V\otimes'W$ sending $(v,w)$ to $(v\otimes w,0)$. And if $h:V\times W\to Z$ is any bilinear map, there exists a linear map $\tilde{h}:V\otimes'W\to Z$ satisfying $h=\tilde{h}\circ\varphi'$, namely let $h_1$ be the unique map $V\otimes W\to Z$ and let $h_2:U\to Z$ be any linear map (say, the zero map) and define $\tilde{h}(x,u)=h_1(x)+h_2(u)$. But there is a compatible isomorphism $k:V\otimes W\to V\otimes'W$ only if $U$ is trivial.

Intuitively, if you don't require uniqueness, there is nothing preventing $V\otimes'W$ from having random elements that are totally unrelated to $V$ and $W$ (e.g., the elements of $U$). The uniqueness requirement says that a linear map out of $V\otimes W$ is determined by where it sends elements of the form $v\otimes w$, which forces those elements to span the entire vector space.

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  • $\begingroup$ In the context of algebraic semantics, uniqueness (in that case of a homomorphism from an initial algebra) corresponds to the "no junk" part of "no junk, no confusion" as your example illustrates in a different context. $\endgroup$ – Derek Elkins left SE Apr 5 '17 at 17:50

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