# Once we've prove the existence, do we really need to prove the uniqueness (of that map) in order to use the universal property?

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.

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.