How can one know $(Vect_k,\otimes_k)$ is not strict monoidal category? Can we identify that given two isomorphic sets are distinct or equal? I can't prove $(Vect_k,\otimes_k)$ is not a strict monoidal category.
I read the answer for $(\mathrm{Vect}_k,\otimes_k)$ as a non-strict monoidal category, but I can't prove that $k\otimes V$ and $V$ are distinct as sets. In the first place, can we identify that given two isomorphic sets are distinct ( or equal ) when we don't know whether they are? Please give me some comments or advice.
 A: In order to check that $V$ and $k\otimes V$ are distinct, you must first give a precise model for $k\otimes V$. There are various ways of constructing a particular model for the tensor product. They are all, of course, isomorphic. But to answer the question is $V=k\otimes V$, one must first define what $k\otimes V$ means as a construction.
It's easier to understand in the context of sets and products. The category $\mathbf {Set}$ is monoidal under the categorical product (which is the cartesian product). However, there is no such thing as the cartesian product of two sets. Therefore, whether $*\times A$ and $A$ are distinct is meaningless. Any particular choice of a product is an arbitrary choice. One common such arbitrary choice is to define $A\times B=\{(a,b)\mid a\in A,b\in B\}$, with the obvious projections. Now it is meaningful to ask whether $*\times A$ and $A$ are distinct. And the answer is that they are as a direct verification shows (unless $A=\emptyset$).
A: From a structuralist point of view, it does not make any sense to ask if two abstract vector spaces (or groups, or sets, etc.) are equal to each other. It is reasonable to ask if two subspaces of a given vector space are equal, but in general the only reasonable way to compare abstract vector spaces is via linear maps, and therefore being isomorphic is the "correct" notion of equality. Some foundations of mathematics have been proposed to even disallow statements like $A=B$ for objects $A,B$ of the same type (ETCS), or make them equivalent to $A \simeq B$ (HoTT).
In material set theories such as ZFC the question if $V = V \otimes K$ holds is well-defined, though. But as Ittay already pointed out, it depends on the specific model we choose for the tensor product, which already indicates that the question does not have much meaning to mathematical practice (where only the universal property of the tensor product is relevant). This will become even more clear with the (absurd) proof below.
Using the standard construction of the tensor product $V \otimes W = F(|V| \times |W|) / B$, where $F(S) := \{f \in K^S : \mathrm{supp}(f) \text{ finite}\}$ is the free vector space on a set $S$, $|V|$ denotes the underlying set of $V$, and $B \subseteq F(S)$ is  generated by the bilinear relations, we have:

There is a trivial(!) vector space $U$ such that $V \otimes W \neq U$ for all vector spaces $V,W$.

In particular, we have $U \otimes K \neq U$ for this space $U$.
Proof: The elements of $|V \otimes W|$ are $f + B$, where $f \in F(S)$. Notice that these elements are sets which have as many elements as $B$. In particular, they are non-empty. Now let $0$ be any set without elements (in ZFC, there is only one such set) and consider the trivial vector space $U=(\{0\},+,\cdot)$. The only element of $|U|=\{0\}$ is empty, thus cannot be equal to any element of a tensor product. $\checkmark$
