Let $K:L^2[0,1]^{d_1}\to L^2[0,1]^{d_2}$ be integral operator $$(Kf)(y) = \int f(x)k(x,y)d x.$$ If $d_1>d_2$ is it possible for $K$ to be injective?, e.g. let's take $d_1=2,d_2=1$.
More generally, does the injectivity of $K$ impose any restrictions on $d_1,d_2$.
Taking $d_1=2$, $d_2=1$. I couldn't immediately find a "trivial" example, but here is one. Let $\{g_n\}_n$ be an orthonormal basis of $L^2[0,1]^2$. Let $\{E_n\}$ be a partition of $[0,1]$ by intervals, and $$ k(x,y)=\sum_n g_n(x)\,1_n(y). $$
For $y\in E_m$, \begin{align} Kf(y) &=\int_0^1\int_0^1f(x_1,x_2)\,g_m(x_1,x_2)\,dx_1\,dx_2. \end{align}
If $Kf=0$ we have, for all $m$, $$\tag1 0=\int_0^1\int_0^1f(x_1,x_2)\,g_m(x_1,x_2)\,dx_1\,dx_2,\ \ \ \ \ \ m\in\mathbb N. $$ As $\{g_m\}$ is total, this implies that $f=0$, and so $K$ is injective.