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Proposition. Let $f\colon X\to Y$ a function. They are equivalent \begin{equation} \begin{split} (i)&\quad f\;\text{is injective};\\ (ii)&\quad\text{For all}\;E\subseteq X, E=f^{-1}\big(f(E)\big)\\ (iii)&\quad\text{For all}\;E\subseteq X, f(X)\setminus f(E)=f(X\setminus E);\\ (iv)&\quad\text{For all}\;\{E_i\}_{i\in I}\subseteq 2^{X},f\bigg(\bigcap_i E_i\bigg)=\bigcap_i f\big(E_i\big) \end{split} \end{equation}

Proof I already proved that $(ii)\iff(i)$ and $(i)\iff(iv)$, then we have that $(ii)\iff(iv)$.

Question. How can I prove that $(ii)\iff(iii)$?

I proved that, in general, $E\subseteq f^{-1}\big(f(E)\big)$ and $f(X)\setminus f(E)\subseteq f(X\setminus E).$

Thanks!

Jack J.
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    See https://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Javi Jan 26 '19 at 11:14

1 Answers1

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Assume iii. For all x, f(X - {x}) = f(X) - {f(x)}.
If x /= y, f(x) = f(y), then f(y) in f(X) - {f(x)}, a contradiction.
Thus iii implies i.

Assume i.
If y in f(X - E), then exists x in X - E with y = f(x). So y in f(X). If y in f(E), then exists a in E with y = f(a).
Thus a = x, a no, no. So y in f(X) - f(E).

If y in f(X) - f(E), then exists x in X with y = f(x).
So x not in E, otherwise y in f(E), another no no.
Hence y in f(X - E).
Thus i iff iii and since ii iff i, ii iff iii.

William Elliot
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