What happens if $(f_1\otimes g_1)(u_1 \otimes v_1) = (f_2 \otimes g_2)(u_2 \otimes v_2)$? Prove that $\lambda$ is injective.
Here is the definition of the linear map $\lambda$:
Let $f: U \to U'$ and $g: V \to V'$ be linear maps. We define their tensor product $f \otimes g: U \otimes V \to U' \otimes V'$ by
$$(f \otimes g)( u \otimes v) = f(u) \otimes g(v)$$ for all $u \in U$ and $v \in V.$ This gives rise to a linear map
$\lambda: \operatorname{Hom}(U, U') \otimes \operatorname{Hom}(V, V') \to \operatorname{Hom}(V \otimes U, U' \otimes V')$ defined by $$(\lambda (f \otimes g))(v \otimes u) = f(u) \otimes g(v).$$
1- what is the name of this map $\lambda$ in literature?
my attempt:
Assume that $$(\lambda (f_1 \otimes g_1))(v_1 \otimes u_1) = (\lambda (f_2 \otimes g_2))(v_2 \otimes u_2) \quad \quad \quad (1)$$ but then what should I get? Should I get that $f_1 \cong f_2, g_1 \cong g_2$ and $v_1 = v_2$ and $u_1 = u_2$? If so how can I fulfill this?
Now, by definition of $\lambda,$ from $(1)$ we get that:
$$f_1(u_1) \otimes g_1(v_1) = f_2(u_2) \otimes g_2(v_2) \quad \quad \quad (2)$$
But then how can I complete this? what property of tensor product I should use to get what I want?
Now, by definition of tensor product of two linear functions given above we will get from $(2)$ that:
$$(f_1\otimes g_1)(u_1 \otimes v_1) = (f_2 \otimes g_2)(u_2 \otimes v_2) \quad \quad \quad (3)$$
But then what?
What if I want to prove that the kernel is zero instead?
 A: One elementary way of seeing this when $U$ and $V$ are finite dimensional is the following. Let $u_i$ be a basis for $U$, and $u_j’$ a basis for $U’$. Then $\mathrm{Hom}(U,U’)$ has basis $E_{ji}$ sending $u_i\mapsto u’_j$, and the other basis vectors for $U$ to zero. Similarly for $\mathrm{Hom}(V,V’)$, having basis $F_{qp}\colon v_p\mapsto v’_q$. Thus $\mathrm{Hom}(U,U’)\otimes\mathrm{Hom}(V,V’)$ has basis $E_{ji}\otimes F_{qp}$.
Next, $U\otimes V$ has basis $u_i\otimes v_p$, and $U’\otimes V’$ has basis $u’_j\otimes v’_q$, and so $\mathrm{Hom}(U\otimes V,U’\otimes V’)$ has basis $G_{jq,ip}$, sending $u_i\otimes v_p\mapsto u’_j\otimes v’_q$.
Finally, $\lambda$ sends $E_{ji}\otimes F_{qp}\mapsto G_{jq,ip}$, and so we have an isomorphism.

Edit.
For infinite dimensional modules the result is more subtle, and the key concept here is that of a Mittag-Leffler module. For a commutative ring $R$, an $R$-module $M$ is Mittag-Leffler if and only if, for every family of $R$-modules $N_p$ the map
$$ M\otimes_R\prod N_p \to \prod(M\otimes_RN_p) $$
is injective.
We claim that every free module is Mittag-Leffler. Take $M=R^{(I)}$ and a family $N_p$. The map
$$ M\otimes\prod N_p \to \prod(M\otimes N_p) $$
sends $m\otimes(n_p)$ to $(m\otimes n_p)$. Suppose we have an element $x$ in the kernel, and write it using a basis $e_i$ for $M$ as
$$ x = \sum_{j\in J}e_j\otimes(n_{p,j}) $$
for some finite subset $J$ of $I$. Then $\sum_je_j\otimes n_{p,j}=0$ for all $p$, so $n_{p,j}=0$ for all $p,j$, so $(n_{p,j})=0$ for all $j$, so $x=0$.
In our situation, when $R=K$ is a field, we can write $U\cong K^{(I)}$, in which case $\mathrm{Hom}(U,U’)\cong\prod_IU’$. Similarly $V\cong K^{(P)}$ and $\mathrm{Hom}(V,V’)\cong\prod_PV’$, as well as $U\otimes V\cong K^{(I\times P)}$ and $\mathrm{Hom}(U\otimes V,U’\otimes V’)\cong\prod_{I\times P}(U’\otimes V’)$.
Now both $U’$ and $\prod_PV’$ are vector spaces, so free $K$-modules, and hence Mittag-Leffler. It follows that
$$ \prod_IU’\otimes\prod_PV’ \rightarrowtail
\prod_I(U’\otimes\prod_PV’) \rightarrowtail
\prod_{I\times P}(U’\otimes V’) $$
as required.
