I don't understand the following statement: Let $k$ be a field, $A$, $B$ and $C$ $k$-algebras, $f:A\otimes_k B\rightarrow C$ a homomorphism of $k$-algebra. To prove that $f$ is one-to-one, it is enough to check that if $\{a_i\}$ are linearly independent in $A$ and $\{b_i\}$ in $B$ then $\{f(a_i\otimes b_j)\}$ are linearly independent in $C$.
I appreciate any help.
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The linear map $f$ sends a basis $\mathcal{B}$ of $A \otimes B$ to a linearly independent set in the image. Therefore $f(\mathcal{B})$ is a basis of the subspace $im(f)$. Any linear map that sends a basis to a basis is an isomorphism. The map $f$ itself is therefore an isomorphism followed by an inclusion.
Suppose that $\{f(a_i \otimes b_j)\}$ are linearly independent in C whenever $\{a_i\}$ and $\{b_j\}$ are linearly independent in $A$ and $B$ respectively. Assume for sake of contradiction that $f$ is not injective. We will produce a set $\{f(a_i \otimes b_j)\}$ which are linearly dependent in $C$ while $\{a_i\}$ and $\{b_j\}$ are independent in $A$ and $B$ respectively.
Because $f$ is not injective, there is a nontrivial element $k$ in the kernel. The kernel is contained in $A \otimes_k B$. So we can express $k$ as a linear combination of elements of the form $a_i \otimes b_j$ where the $a_i$ are part of a basis for $A$ and the $b_j$ are part of a basis for $B$. In particular, they are linearly independent.
So we can write $k = \sum_{i,j} c_{i,j} (a_i \otimes b_j)$, where this is a finite sum and at least one of the $c_{i,j}$ is nonzero. Then $f(k) = 0$, so $\sum_{i,j} c_{i,j} f(a_i \otimes_k b_j) = 0$ because $f$ is a homomorphism. This implies $\{f(a_i \otimes_k b_j)\}$ are not linearly independent while $\{a_i\}$ and $\{b_j\}$ are. A contradiction.