A friend od mine stated that
$0 \rightarrow \Bbb Z \xrightarrow{i_1} \Bbb Z \oplus \Bbb Z/2\Bbb Z\xrightarrow{\text{pr}_2} \Bbb Z/2\Bbb Z \rightarrow 0 \tag{*} $
with $i_1$ the inclusion in the first component and $\text{pr}_2$ the projection in the second component,
was a short exact sequence (s.e.s). Is this true?
To begin with, I am not completely sure I understand what the elements of the direct sum are and how the inclusion works:
I guess $\Bbb Z \oplus \Bbb Z/2\Bbb Z=\{(n,[0]), n \in \Bbb Z\}\cup \{(n,[1]),n \in \Bbb Z\}$
And what is the identity ? I guess it's $(0, [0])$
-Exactness at $\Bbb Z/2\Bbb Z$ is clear because the projection is always surjective
-For exactness at $\Bbb Z$ we need $\{0\}$=$\text {Ker}(\text{i}_1)$.
-In order to have exactness at the direct sum term we need $\text {Im}(i_1)$=$\text {Ker}(\text{pr}_2)$.
$\text {Ker}(\text{pr}_2)=\{(n,[0]), n \in \Bbb Z\}$, because these are the elements that are map to [0] in $\Bbb Z/2\Bbb Z$ right?
$\text {Im}(i_1)=?$ I am not sure how this inclusion works, if each integer $n$ is sent to both $(n,[0])$ and $(n,[1])$ that would not be a function.
Could you help me clarify if my guesses are correct and what are then the correct $\text {Ker}(\text{pr}_2)$ and $ \text {Im}(i_1)$ ? This should answer whether $(*)$ is a short exact sequence or not
Although it looks like the question was basically answered in the comments, I will write it down for the sake of the question not showing as unanswered.
For any two abelian groups (or modules) $A$ and $B$ sequence $$0 \rightarrow A \xrightarrow{i_A} A \oplus B \xrightarrow{p_B} B \rightarrow 0$$ is short exact with the maps being defined as $i_A(a) = (a,0)$ and $p_B(a,b) = b$.
To see this note that $(p_B\circ i_A)(a) = p_B(a,0) = 0$, so the composition is the zero map. Furthermore \begin{align}\ker p_B &= \{ (a,b)\in A\oplus B\mid p_B(a,b) = 0 \}\\ &= \{ (a,b)\in A\oplus B\mid b = 0 \} = \{ (a,0)\mid a\in A \} = \operatorname{im} i_A.\end{align} Exactness at $A$ and $B$ follows from injectivity of $i_A$ and surjectivity of $p_B$.
Let's answer your specific questions.
The direct sum you have is $\mathbb Z\oplus\mathbb Z/2\mathbb Z = \{ (m,[n])\mid m,n\in\mathbb Z\}$ which you can write as disjoint union $\{(n,[0])\mid n \in \Bbb Z\}\cup \{(n,[1])\mid n \in \Bbb Z\}$ if you want.
Identity in any direct sum $A\oplus B$ is always given as $(0_A,0_B)$ where $0_A$ is the identity of $A$ and $0_B$ is the identity of $B$. This follows from definition of addition in $A\oplus B$ as $(a_1,b_1)+(a_2,b_2) = (a_1+a_2,b_1+b_2)$.
What you wrote before looking at the image of your map $i_1$ is all correct. Here you seem to be confused how $i_1$ is defined, it's simply $i_1(n) = (n,[0])$, elements $(n,[1])$ are not hit by $i_1$ for any $n$. Nor they need to be, $i_1$ doesn't need to be surjective (and won't be in general). Therefore, the image of your $i_1$ is $\{(n,[0])\mid n\in\mathbb Z\}$ which is exactly the kernel your wrote.