Universal property of the quotient ring

Statement

Let $R$ be a ring, let $I$ be a two-sided ideal of $R$ and let $\pi \colon R \to R/I$ be the quotient map. Let $f \colon R \to S$ be a ring homomorphism with $I \subseteq \ker f$. Then there is exactly one ring homomorphism $\bar{f} \colon R/I \to S$ such that $\bar{f} \circ \pi = f$. It is given by $\bar{f}(a + I) = f(a)$.

Remarks

The map $\bar{f}$ is well defined: if $a + I = b + I$, then $a - b \in I \subseteq \ker f$, so $f(a) = f(b)$. The image of $\bar{f}$ is the image of $f$, and $\ker \bar{f} = (\ker f)/I$. With $I = \ker f$, this gives the first isomorphism theorem.

Explanations

L1 human

A homomorphism that sends all of $I$ to zero does not see the difference between $a$ and $a + x$ for $x \in I$. So it already makes sense on the cosets of $I$. This gives a homomorphism from the quotient ring.