I have been trying to understand closed immersions and reduced induced subscheme structure and I have asked questions, and the answers have been very helpful. I have one more thing that I find confusing about this, and any explanation is appreciated.
The set up is as follows (this is the set up as in the answer to Question about given $f: X \to Y$ a morphism of schemes, factoring it as $X \to \overline{f(X)} \to Y$): We have a morphism of schemes $f: X \to Y$, $X, Y$ reduced. $U_1, U_2$ are affine open of $X$ and $f(U_i) \subseteq V_i$, where $V_i$ are affine open of $Y$. We put a reduced induced subscheme structure on $\overline{f(X)} \subseteq Y.$ So we have $$U_1 \xrightarrow{g_1} \overline{f(X)} \cap V_1 \xrightarrow{i_1} V_1$$ and $$U_2 \xrightarrow{g_2} \overline{f(X)} \cap V_2 \xrightarrow{i_2} V_2.$$
Then we have for $U_{12} \subseteq U_1 \cap U_2$, open affine and $V_{12} \subseteq V_1 \cap V_2$, open affine with $f(U_{12}) \subseteq V_{12}$, two sequences $$U_{12} \xrightarrow{g_{12}} \overline{f(X)} \cap V_{12} \xrightarrow{i_{12}} V_{12}$$ and $$U_{12} \xrightarrow{g_{21}} \overline{f(X)} \cap V_{12} \xrightarrow{i_{21}} V_{12}.$$
Here $g_{12}$ is the restriction of $g_1$ on $U_{12}$ and $g_{21}$ is the restriction of $g_2$ on $U_{12}$. $i_{12}$ is the restriction of $i_1$ on $\overline{f(X)} \cap V_{12}$ and similarly for $i_{21}$.
What I am trying to understand is how do we know that $i_{12} = i_{21}$? as in how come they are the same map? In the two sequences I know that $\overline{f(X)} \cap V_{12}$ have the same reduced induced structure as a closed subset of $V_{12}$ (this I understand now thanks to A reduced induced closed subscheme structure of an affine scheme), but how does one deduce that the choice of closed immersion is unique? so that I can conclude $i_{12} = i_{21}$.. Thank you.