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Thm: Let $(x,τ)$ be a fuzzy topological space. The identity function $f:(X,τ)→(X,τ)$ is fuzzy continuous.
Proof: Let $A∈τ$. Since $f^{-1}(A)=A(f)=A∘f=A$, then $f^{-1} (A)∈τ$.

The same method is used in

Thm: A composition of fuzzy continuous functions is fuzzy continuous. Let

Proof: Let $f:(X,\tau_1)\rightarrow (Y,\tau_2)$,$g:(Y,\tau_2)\rightarrow (Z,\tau_3)$ fuzzy continuous functions. Let $A∈τ_3$, then $(g∘f)^{-1} (A)=A∘(g∘f)=(A∘g)∘f=f^{-1}(A∘g)=f^{-1} (g^{-1}(A)) ⇒ (g∘f)^{-1} (A)∈τ_1 ⇒g∘f$ is fuzzy continuous.

Why is $(g∘f)^{-1} (A)=A∘(g∘f)$? Is this always true? i.e. for any two function $f,g$, then $f^{-1}(g)=g(f)$.

CyclotomicField
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Isaac
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  • The inverse of the identity is the identity. It is true if $f^{-1}=f$ – Marco Mar 14 '25 at 14:50
  • @Marco, but the same method is used in the second theorem, where neither $f$ or $g$ are identity functions – Isaac Mar 14 '25 at 14:59
  • How is $A(f)$ defined? – CyclotomicField Mar 14 '25 at 15:32
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    What is the reference? This question needs that context, unless this is all well-known to fuzzy-set people. My guess: the equality $g^{-1}(A)=A\circ g$ is true by definition, since if $A$ is a fuzzy set in the sense of a function $Z\to I$, then this is the natural way to define inverse image. (Note: $g^{-1}$ is not the inverse, but the inverse image.) – Kyle Miller Mar 14 '25 at 15:53
  • @CyclotomicField, Fuzzy continuous if the inverse image under $f$ of any open fuzzy set in $(Y,\tau_2)$ is an open fuzzy set in $(X,\tau_2)$; i.e. $f^{-1}(A)\in \tau_2$ whenever $A\in \tau_1$. – Isaac Mar 14 '25 at 17:25
  • @Isaac if $A$ is a fuzzy set then how are you composing it with a function? What's the definition of $A(f)$ and $A \circ f$? – CyclotomicField Mar 14 '25 at 18:07
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    @CyclotomicField, A fuzzy set $A$ in $X$ is a function defined from $X$ to $I$. – Isaac Mar 14 '25 at 18:08

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We can see that $f^{-1}(g) \neq g(f)$ in general if we consider $f(x)=2x$ and $g(x)=x+1$.

What is true in general is that $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$. We can see this because $$(f^{-1} \circ g^{-1})\circ (g \circ f) = f^{-1}\circ (g^{-1} \circ g) \circ f = f^{-1}\circ f = \text{Id}$$ with $\text{Id}$ the identity function. So associativity is used in the proof but it's unclear to me what they intended in that section. I suspect it may be errata.

CyclotomicField
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