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Let $f \colon M \to N$ a function. Show that the following affirmations are equivalent:

(a) $f$ is injective

(b) every two subsets $M_1$ and $M_2$ of $M$, $f(M_1 \cap M_2)=f(M_1) \cap f(M2)$

(a) $\implies$ (b):

Let $y \in f(M_1 \cap M_2)$

$\implies$ exists $x$ an element from $M_1 \cap M_2$ such that $f(x)=y$

$\implies$ exists $x \in M_1$ such that $f(x)=y$ and exists $x \in M_2$ such that $f(x)=y$

$\implies y \in f(M1) \cap f(M2)$.

So, $f(M_1 \cap M_2)$ is a subset of $f(M_1) \cap f(M_2)$.

Let $y \in f(M_1) \cap f(M_2)$. How can I prove, using the injectivity of the function $f$,that $y \in f(M_1 \cap M_2)$?

(b)$\implies$ (a)

$f(M_1 \cap M_2)=f(M_1) \cap f(M_2)$ means that every $y$ from $f(M_1 \cap M_2)$ belongs to $f(M_1) \cap f(M_2)$ and every $y$ from $f(M_1) \cap f(M_2)$ belongs to $f(M_1 \cap M_2)$.

How can I prove that $f$ is injective?

Community
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Adrian Berteanu
  • 49

2 Answers2

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Suppose $f$ injective. Notice that $f(M_1\cap M_2)\subset f(M_1)\cap f(M_2)$ always. So let $y\in f(M_1)\cap f(M_2)$. Then, there is a $x$ and a $z$ s.t. $y=f(x)=f(z)$ and thus $x=z$ by injectivity. Therefore $x\in M_1\cap M_2$ and thus $y\in f(M_1\cap M_2)$.

I suppose we are in a Hausdorf space (otherwise, I don't think that the result is correct). Conversely, suppose $f(M_1\cap M_2)=f(M_1)\cap f(M_2)$ for all $M_1,M_2$. If $f$ not injective, there is $x\neq y$ with $f(x)=f(y)$. In particular, we can take $U,V$ s.t. $x\in U$, $y\in V$ and $U\cap V=\emptyset$. Then $f(U)\cap f(V)\neq \emptyset$ and $f(U\cap V)=\emptyset$. This is a contradiction.

Rick
  • 1,707
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Taking $m_1\in M_1$ and $m_2\in M_2$ and the set: $$ X=\{m_1\}\cap\{m_2\}$$ such that $f(m_1)=f(m_2)=f(m)$. So, by the given property: $$ \begin{split} f(X) & =\{f(m_1)\}\cap\{f(m_2)\}\\ &=\{f(m)\} \\ & \neq \emptyset \end{split}$$ So: $$ X\neq \emptyset$$ and so: $$ m_1=m_2$$ which, by definition, makes $f$ injective.