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In a category, if an idempotent $f:a \longrightarrow a$ splits, then any two splittings are isomorphic. Let $i: b \longrightarrow a$ and $p: a \longrightarrow b$ be such that $i\circ p=f$ and $p\circ i=id_b$ and also $j:c \longrightarrow a$ and $q: a\longrightarrow c$ be such that $j\circ q=f$ and $q\circ j=id_c$. Then $b$ is isomorphic to $c$. What is the isomorphism between $b$ and $c$?

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2 Answers 2

There is only one distinguished morphism $b \to c$, namely $qi$, and the same for $c \to b$, namely $pj$. One checks without effort that they are inverse to each other. For example, we have $qi \, pj = qfj = qjqj=1$.

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As Martin said: The morphism $g=qi:b\to c$ is the inverse of $h=pj:c\to b$. I'd like to add a different approach:
We see that $p\circ 1_a=1_bp=(pi)p=p(ip)=p\circ f$ is a cocone over the diagram $a\xrightarrow{1}a\xrightarrow{f}a$.
Now let's assume that $q:a\to c$ is another cocone (we don't need the $j$ for now). Then $g=qi:b\to c$ is an arrow such that $gp=qip=qf=q1_a=q$, so $g$ is a morphism between cocones. Since $p$ is epic, it is the only such morphism, which means that $p:a\to b$ is the colimit of the diagram $a\xrightarrow{1}a\xrightarrow{f}a$.
If we assume the existence of a $j$ as in your question, then $c$ is another colimit, so there must be an isomorphism $b\simeq c$.

Also note that since $p^\text{op}i^\text{op}=f^\text{op}$, $\ i^\text{op}p^\text{op}=1_b$, and $f^\text{op}f^\text{op}=f^\text{op}$, it follows dually that $i:b\to a$ is the limit of the diagram.

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