Why does exactness of $X \to Z \to Y \to 0$ imply $\operatorname{im} p=Y$?

Conditions

• $X \xrightarrow{i} Z \xrightarrow{p} Y \to 0$ is an exact sequence.
• Exactness is defined as $\operatorname{im}(\text{incoming}) = \ker(\text{outgoing})$ at each internal object.

Reasoning, step by step

1. Identify the map leaving $Y$ in the sequence, which is the zero map $Y\to 0$.
2. Recall the definition of exactness at $Y$: $\operatorname{im}(p) = \ker(Y\to 0)$.
3. Determine the kernel of the zero map $Y\to 0$. Since every element in $Y$ maps to $0$ in the zero module, the kernel is all of $Y$.
4. Conclude that $\operatorname{im}(p) = Y$.
5. This equality is used in the proof of the cokernel lemma to replace $\operatorname{im}(p)$ with $Y$.
6. The video explicitly states this step: 'Now use the fact that the sequence ends with $Y\to 0$. Exactness at $Y$ forces $\operatorname{im} p = \ker(Y\to 0)=Y$, so $p$ is surjective.'

Example

The video states: 'Now use the fact that the sequence ends with $Y\to 0$. Exactness at $Y$ forces $\operatorname{im} p = \ker(Y\to 0)=Y$, so $p$ is surjective. Substituting this into the previous line upgrades the statement to $Z/\ker p \simeq Y$.'

Common misconceptions

• Thinking that $\operatorname{im}(p)$ is always $Y$ for any map $p:Z\to Y$. It is only true if $p$ is surjective, which is guaranteed here by exactness at $Y$.
• Confusing the kernel of $Y\to 0$ with the image of $Y\to 0$. The image is $0$, but the kernel is $Y$.
• Forgetting that exactness applies at every object in the sequence, including the rightmost non-zero object $Y$.

Watch the explanation

Explore next

• Why does exactness of $X \to Z \to Y \to 0$ imply $\operatorname{im} p=Y$?
• What does it mean for a sequence to be exact at $Y$?
• How is surjectivity related to exactness?