Prove the following generalization of the Factorization Theorem Prove the following generalization of the Factorization Theorem: Let $f_1$ and $f_2$ be homomorphisms from the ring $R$ onto the rings $R_1$ and $R_2$, respectively.If       $\ker f_1 \subseteq \ker f_2$, then there exists the unique homomorphisms $\bar{f}: R_1 \to R_2$ satisfying $f_2= \bar{f} \circ f_1$. For any element $f_1(a) \in R_1$ , define $\bar{f} (f_1(a))= f_2(a)$.
 A: It is a general fact that, given sets $S_1, S_2, S_3$ and functions $f:S_1\rightarrow S_2$, $g_1, g_2:S_2\rightarrow S_3$ and $h:S_1\rightarrow S_3$, such that $h = g_1\circ f = g_2\circ f$, if $f$ is surjective, then $g_1 = g_2$. This fact assures that, if a homomorphism $\overline{f}$ satisfying $f_2 = \overline{f}\circ f_1$ exists, it is unique. It remains to ascertain the existence of such a homomorphism.


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*The First Ring Isomorphism Theorem assures us that $x\mapsto f_1^{-1}\left(\{x\}\right)$ is an isomorphism from $S_1$ to $R/\mathrm{ker}(f_1)$.

*The Third Ring Isomorphism Theorem assures us that the quotient ring $\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}$ is well-defined, and that the mapping $[x]_{\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}}\mapsto\cup[x]_{\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}}$ constitutes an isomorphism from $\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}$ to $R/\mathrm{ker}(f_2)$. Since the Canonical Map $x\mapsto[x]_{\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}}$ is a homomorphism from $R/\mathrm{ker}(f_1)$ to $\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}$, we obtain that the mapping $x\mapsto \cup [x]_{\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}}$ is a homomorphism from $R/\mathrm{ker}(f_1)$ to $R/\mathrm{ker}(f_2)$.

*The First Ring Isomorphism Theorem assures us that the mapping $[x]_{R/\mathrm{ker}(f_2)}\mapsto f_2(x)$ from $R/\mathrm{ker}(f_2)$ to $R_2$ is well-defined (i.e. independent of the representative $x$), and induces an isomorphism from $R/\mathrm{ker}(f_2)$ to $S_2$.
Stringing the three homomorphisms together, we obtain the homomorphism
$$
\overline{f}:S_1\rightarrow S_2,\hspace{1cm}\overline{f}(a) := f_2\left(\bigcup\left[f^{-1}\left(\{a\}\right)\right]_{\frac{R/\mathrm{ker}(f_1)}{R/\mathrm{ker}(f_2)}}\right),
$$
which satisfies the equation $f_2 = \overline{f}\circ f_1$.
