How to show if a function is idempotent?

Question

Let $f:B\to B$ be a function, with $B'\subseteq B$ the range of $f$. Show that $f$ is idempotent if $b' \in B$ is a fixed point of $f$? Show that $f$ is an idempotent function!

I am having trouble writing a clear proof for this question. I looked up definitions online and found out that a function is idempotent iff $f \circ f = f$. In this question, we have the identity function: $$(1_{B'} \circ 1_{B})(b') \implies 1_{B'} (1_{B}(b'))\implies 1_{B'}(b')$$ I am stuck after this point. I am not sure about the correct answer to this question. Please help me out! Please suggest how to write a correct proof for this question. What am I missing in my explanation? Thanks!

Answer 1

First of all, you've translated or copied the question incorrectly. Precise phrasing is important.

Let $f:B \to B$, and let $B' \subseteq B$ denote the range of $f$. Show that $f$ is idempotent if every $b' \in B'$ is a fixed point of $f$.

Hint: It's important to understand the definitions at work here

• We say that a function $f:B \to B$ is idempotent if $f \circ f = f$. That is, for every $b \in B$, we have $f(f(b)) = f(b)$.
• We say that $x$ is a fixed point of $f$ if $f(x) = x$.
• We say that an element $b'$ is in the range of $f$ if there exists a $b$ for which $b' = f(b)$.

If you really understand the definitions, it should be easy to see how they can interact. In particular: if $f(b)$ is a fixed point of $f$, what can we say about $f(f(b))$?

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Here's a correct proof:

Suppose that every point of the image of $f$ is a fixed point of $f$. That is, for every $b$ in $B$, the element $f(b)$ is a fixed point of $f$. So, for every $b$ in $B$, we have $f(f(b)) = f(b)$, which is to say that for every $b \in B$ $[f \circ f](b) = f(b)$. Thus, we conclude that $f \circ f = f$. That is, $f$ must be idempotent.