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Let $f\colon A \to B$ and $C \subset A$. Define $f[C]=\{b∈B : b=f(a) \text{ for some } a \in C\}$. Prove or disprove each of the statements:

a) If $f$ is a bijection, then $f[A\setminus C] = B\setminus f[C]$.

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Let $f$ be a bijection. since $f$ is onto then $f[A]=B$ and since $f$ is 1 to 1 then $f[A\setminus C] = B\setminus f[C]$ since every $A$ correspond to a $B$ and every $C$ correspond to some $f[C]$ so you remove that part from $B$.

Is this correct?I’m not sure if this is enough or all i need. Is there a more concise proof?

b)If $f[A\setminus C] = B\setminus f[C]$ for every $C \subset A$, then $f$ is a bijection

I’m not sure how to do this one.

do I say that do I say this need to be 1 to 1 so $A$ doesn’t make a $f[C]$. Any help would be great. Thanks!

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For part (a), you need to prove subset inclusion in both directions; that is, $f(A\diagdown C)\subseteq B\diagdown f(C)$ and $f(A\diagdown C)\supseteq B\diagdown f(C)$. Here’s how to write it up in full detail using proper mathematical language.

$(\subseteq)$ Let $y\in f(A\diagdown C)$. Then $y=f(a)$ for some $a\in A\diagdown C$. We know $y\in B$ because $f:A\rightarrow B$. If $y\in f(C)$, then $y=f(c)$ for some $c\in C$. But then $f(a)=f(c)$ implies $a=c$ since $f$ is injective, contradicting that $a\notin C$. Hence $y\in B\diagdown f(C)$.

$(\supseteq)$ Let $b\in B\diagdown f(C)$. Since $f$ is surjective, $b=f(a)$ for some $a\in A$. If $a\in C$, then $b=f(a)\in f(C)$ is contradictory. So $a\in A\diagdown C$, and hence $b=f(a)\in f(A\diagdown C)$.

For part (b):

Take $C=\varnothing$, the empty set. Then $f(A)=f(A\diagdown \varnothing)=B\diagdown f(\varnothing)=B$, so $f$ is surjective.

Suppose $f(a)=f(a’)$ for some $a,a’\in A$ with $a\neq a’$. Taking $C=\{a’\}$, we see that $f(A\diagdown \{a’\})=B\diagdown \{f(a’)\}$. Since $a\in A\diagdown \{a’\}$, we have $f(a)\in f(A\diagdown \{a’\})=B\diagdown \{f(a’)\}$, which implies $f(a)\neq f(a’)$, a contradiction. So we must have $a=a’$, therefore $f$ is injective.

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