How to prove that $f\otimes g: V\otimes W\to X\otimes Y$ is a monomorphism Let $V, W, X, Y$ be the vector spaces over a field $\mathbb K$. Suppose that $f: V\to X$ and $g: W\to Y$ are the homomorphism of $\mathbb K$-vector spaces.
How to prove that $f\otimes_{\mathbb K} g: V\otimes_{\mathbb K} W\to X\otimes_{\mathbb K} Y$ is a monomorphism map if and only if $f$ and $g$ are also monomorphism maps?
Thanks all for help!
 A: A basic structure theorem for vector spaces is that given any linear map
$$ T : A \to B $$
you can decompose everything into direct sums:
$$ A \cong \ker(T) \oplus \operatorname{coim}(T) \qquad \qquad B \cong \operatorname{im}(T) \oplus \operatorname{coker}(T) $$
such that $T$ restricts to the canonical isomorphism $\operatorname{coim}(T) \to \operatorname{im}(T)$.

You can use the fact tensor products distribute over direct sums to decompose your problem into smaller pieces that are easy to analyze.
A: The key to your problem are following two lemmas

Lemma 1. Any vector $z\in V\otimes W$ can be writtes as
  $$z=\sum_i v_i\otimes w_i,$$
  where $v_1,\dots,v_k\in V$ are lineary independent and $w_1,\dots,w_k\in W$ are arbitrary.
Lemma 2. Let $x_1,\dots,x_p\in X$ be arbitrary vectors and let $y_1,\dots,y_p\in Y$ be lineary independent vectors. We have that
  $$\sum_j x_j\otimes y_j=0\implies\forall_j x_j=0.$$

Lets go back to your situation. Assume first that $f,g$ are mono and take $z\in V\otimes W$. According to Lemma 1 we can assume that
$$z=\sum_i v_i\otimes w_i,$$
where $v_1,\dots,v_k\in V$ are lineary independent and $w_1,\dots,w_k\in W$ are arbitrary. Suppose that $(f\otimes g)(z)=0$. Thus
$$(f\otimes g)(z)=(f\otimes g)\left(\sum_i v_i\otimes w_i\right)=\sum_i f(v_i)\otimes g(w_i)=0.$$
Now consider a linear subspace $W_0$ of $W$ spanned by $w_1,\dots,w_k$ and put $Y_0 = g(W_0)$. Fixing a base $y_1,\dots,y_p\in Y_0$ we get that
$$g(w_i) = \sum_j a_i^j y_j$$
for some numbers $a_i^j\in \mathbb{K}$. Thus
$$\sum_i f(v_i)\otimes g(w_i) = \sum_if(v_i)\otimes (\sum_j a_i^j y_j) = \sum_j \left(\sum_i a_i^j  f(v_i)\right) \otimes y_j$$
But since $\sum_i f(v_i)\otimes g(w_i)=0$ and $y_1,\dots y_p$ are lineary independent, form the lemma 2 we get that
$$\sum_i a_i^j f(v_i) = 0$$
for every $j=1,\dots,p$. But $f$ is linear, so we have that
$$f\left(\sum_i a_i^j v_i\right)=0.$$
And since $f$ is mono we have that $\sum_i a_i^j v_i=0$ and from linear independency of $v_1,\dots, v_k$ we get that all $a_i^j=0$. Recall that
$$g(w_i) = \sum_j a_i^j y_j=0$$
and since $g$ is mono we have that $w_i=0$ for all $i$. Thus $z=0$.
Assume now that $f\otimes g$ is mono and suppose that $f(v)=0$. Take any non-zero $w\in W$. Then
$$(f\otimes g)(v\otimes w) = f(v)\otimes g(w) = 0.$$
$f\otimes g$ were mono, so $v\otimes w=0$. But we assumed that $w\neq 0$ so $v=0$ (notice that this is a special case of Lemma 2). Thus $f$ is mono. Analogously we show that $g$ is mono.
Lemma 1 and Lemma 2 may be found in Greubs "Multilinear Algebra" on the page number 7.
