Let $f:X\to Y$ be a morphism of schemes and let $Z\subset Y$ be its scheme-theoretic image. If $T$ is any other scheme, consider the induced morphism $g=f\times 1_T:X\times T\to Y\times T$.
Question. Is the scheme theoretic image of $g$ equal to $Z\times T$?
Let $j:Z\times T\to Y\times T$ be the natural closed immersion. What I need to check is that the kernel of $$p:\mathcal O_{Y\times T}\to g_\ast\mathcal O_{X\times T},$$ which is the ideal of the scheme-theoretic image of $g$, agrees with the kernel of $$q:\mathcal O_{Y\times T}\to j_\ast\mathcal O_{Z\times T},$$ which is the ideal defining $Z\times T\subset Y\times T$.
To simplify matters, let us assume $X$ is reduced or noetherian, so that the scheme-theoretic image can be computed affine-locally. So, given an affine open subset $R=\textrm{Spec }B\subset Y\times T$, I am left with checking that the kernel of $$B\to \Gamma\bigl(R,g_\ast\mathcal O_{X\times T}\bigr)=\mathcal O_{X\times T}\bigl(g^{-1}(R)\bigr)$$ agrees with the kernel of $$B\to \Gamma\bigl(R\cap (Z\times T),j_\ast \mathcal O_{Z\times T}\bigr)=\mathcal O_{Z\times T}\bigl(R\cap (Z\times T)\bigr).$$ Now it is probably the time to use that $Z$ is the scheme-theoretic image of $f$, but I do not see exactly how this would help. Does anyone have any suggestion? Thanks in advance!