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I couldn't recall the UPQ from my memory exactly, so I wrote the following:

Let $R, S$ be rings, $I$ be an ideal of $R$, $\pi:R\to R/I$ be the canonical quotient ring homomorphism and $\varphi:R\to S$ be a surjective ring homomorphism. Then there exists isomorphism $\overline{\varphi}:R/I\to S$, such that ${\varphi}=\overline{\varphi}\circ \pi$.

I'm looking at the original statement, but it is phrased differently - there is no word "surjective" and no word "isomorphism", and $I$ is said to be subset of $\ker\varphi$. But could there possibly be some truth to what I wrote?

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    $\begingroup$ I believe you mean $\varphi = \overline{\varphi} \circ \pi$, right? This equation alone tells you that $I \subseteq \ker(\varphi)$ since $I = \ker(\pi)$. You might also want to think about the case $S = R$ and $\varphi = \text{id}_R$. $\endgroup$ – Matthias Klupsch Apr 13 '17 at 10:08
  • $\begingroup$ @MatthiasKlupsch Yes, I made a typo here, will fix it now. $\endgroup$ – sequence Apr 13 '17 at 19:33
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The general universal property of quotients states:

For a ring homomorphism $\varphi:R\to S$, and and ideal $I\subseteq\ker\varphi$, there exists a unique homomorphism $\overline{\varphi}:R/I\to S$ such that $\bar{\varphi}\circ\pi=\varphi$, where $\pi:R\to R/I$ is the usual projection.

In the case where $\varphi$ is surjective, $\overline{\varphi}$ must be surjective as well, since given any $s\in S$, there exists an element $r\in R$ such that $s = \varphi(r) = \overline{\varphi}(r+I)$.

The last bit should be a little more precise; you should have that $I=\ker\varphi$, not just containment $\subseteq$. When $I=\ker\varphi$, the first isomorphism theorem implies that the induced map $\overline{\varphi}:R/I\to S$ is an isomorphism. So, there is some truth to what you wrote, as long as you specify that $I=\ker\varphi$.

As an example, let $\varphi:\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$ be the projection. Note that this is surjective. Then, we have $4\mathbb{Z}\subseteq\ker\varphi = 2\mathbb{Z}$, and thus there exists a unique map $$\overline{\varphi}:\mathbb{Z}/4\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}$$ such that $\overline{\varphi}\circ\pi=\varphi$. However, $\overline{\varphi}$ is clearly not an isomorphism, since one ring has $2$ elements while the other has $4$. So the condition $I=\ker\varphi$ is necessary to conclude that the induced map is an isomorphism.

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