Question regarding the second isomorphism theorem proof

Question

I have a question regarding a part of the proof of the second isomorphism theorem.

Let $I_1, I_2$ be ideals of a ring $R$. Then it's clear that the function $f: I_2 \rightarrow I_1 + I_2 \ / \ I_1$ such that $f(y_2) = \overline{0+y_2}$ is well defined (this is, $f(y_2) \in I_1 + I_2 \ / \ I_1$

I can't see why it is clear that $ \overline{0+y_2} \in I_1 + I_2 \ / \ I_1$.

Answer 1

$\overline{0+y_2}=\overline{\vphantom{0}y_2}=y_2+I_1$.

A general element of $(I_1+I_2)/I_1$ is the set of elements of the form $\;y_1+y_2+I_1=y_2+(y_1+I_1)$ ($y_1\in I_1,\:y_2\in I_2$). Now just observe that for any $y_1\in I_1$, one has $\;y_1+I_1=I_1$.

Answer 2

I assuming that $$(I_1+I_2)/I_1=\{ \overline{y_1+y_2} \mid y_1 \in I_1, y_2 \in I_2\}$$ so since $0 \in I_1$ for every $y_2 \in I_2$ we have $\overline{0+y_2} \in (I_1+I_2)/I_1$.