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Proposition:

Suppose $A ≠ \emptyset$ and $f: A \rightarrow A$. Suppose that for all $g: A \rightarrow A$, $f \circ g = f$. Prove that $f$ is a constant function.

My attempt:

Definitions:

  1. If $f$ is a constant function, then exists some $a \in A$ such that

$$\forall x \in A (f(x) = a)$$

  1. Composition of functions $$f \circ g = \{(a,b) \in A \times A \mid \exists c \in A \bigl((a,c) \in g \land (c,b) \in f\bigr)\}$$

By contrapositive.

Suppose $f$ is not a constant function.

Suppose $g$ is a constant function.

Then exists some $a \in A$, such that for arbitrary $x \in A$, $(x,a) \in g$.

Since $a \in A$, exists some $y \in A$ such that $(a,y) \in f$.

Since $f$ is not a constant function, there exist some $k_1,k_2 \in A$ such that $(k_1,k_2) \in f$ and $y ≠ k_2$.

Since $k_1 \in A$, we know that $(k_1,a) \in g$ and therefore $(k_1,y) \in f \circ g $.

$f \circ g$ is a function, thus we can conclude that $(k_1,k_2) \notin f \circ g$

Hence $f ≠ f \circ g$. $\Box$

Is it correct?


My initial plan was to prove it directly, but after struggling for hours, I failed to find any way to do it. Can someone show me how to prove the proposition directly?

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We may assume that $A$ has at least two elements (if $|A|=1$, then we have only the identity map, that it's constant in this case).

Suppose $f$ is not a constant function. We want to prove there exists a $g:A \rightarrow A$ such that $f \circ g \neq f$.

Since $f$ is not constant, there exist some $a,b \in A$, $a \neq b$ such that $f(a) \neq f(b)$. Take $g$ to be $g(x):=b$ for every $x \in A$. Now $(f \circ g) (a)=f(b)$. Hence we have $(f \circ g) (a)=f(b) \neq f(a)$, so $f \circ g \neq f$.

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A direct proof: Since $A$ is non-empty, we can pick an element $a \in A$. Take a constant function $g:A\to A$ such that $g(x) = a$ for all $x\in A$. Then $f(g(x)) = f(a)$ for any $x \in A$, so that $f \circ g$ is a constant function, thus $f$ is also a constant function.

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You only need to prove that for any $x,y \in A$, $f(x) = f(y)$ holds. Let $x,y \in A$. Choose $g$ such that $g(x) = g(y)$, e.g. $g(z) = g(x)$ for all $z \in A$. Then $f(x) = f(g(x)) = f(g(y)) = f(y)$.

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