$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

Let $x\in A$, then $f\upharpoonright A\left(x\right)=f\left(x\right)$, and $(f\circ \iota )\left(x\right)=f\left(\iota \left(x\right)\right)=f\left(x\right)$ thus the two functions are equal.

Let $x\in f\left(A\right)$ so that $x=f\left(a\right)$ where $a\in A$ we then know that $a\in B$ so therefore $x\in f\left(B\right)$

$p\in f(\bigcup ℳ)$ iff $p=f\left(x\right)$ where $x\in \bigcup ℳ$ iff there exists some $M\in ℳ$ such that $x\in M$ which shows that $f\left(x\right)\in f\left(M\right)$ iff $p\in \bigcup \{f(M):M\in ℳ\}$ as needed.

Suppose that $p\in f(\bigcap ℳ)$ iff $p=f\left(x\right)$ where $x\in \bigcap ℳ$ iff $\forall M\in ℳ$, $x\in M$ therefore $f\left(x\right)\in f\left(M\right)$ so that $p=f\left(x\right)\in \bigcap \{f\left(M\right):M\in ℳ\}$

We already know that $\subseteq$ holds true therefore we do the other direction so suppose that $p\in \bigcap \{f\left(M\right):M\in ℳ\}$ therefore $p\in f\left(M\right)$ for all $M\in ℳ$ so that there is some $x_M\in M$ such that $p=f\left(x_M\right)$. We note that for all $J,K\in M$ we know that $f\left(x_J\right)=p=f\left(x_K\right)$ so that by injectivity we have $x_J=x_K$ let $x^{\prime }$ be their common value and note that $x^{\prime }\in \bigcap ℳ$ and that $p=f\left(x^{\prime }\right)$ so that $p\in f(\bigcap ℳ)$

Let $p\in f\left(A\right)\setminus f\left(B\right)$ therefore $p\in f\left(A\right)$ so that $p=f\left(x\right),x\in A$ and $p=f\left(x\right)\notin f\left(B\right)$, therefore it must be the case that $x\notin B$, therefore $x\in A\setminus B$ so we can conclude that $f\left(x\right)=p\in f(A\setminus B)$ as needed.

We already know that $f\left(A\right)\setminus f\left(B\right)\subseteq f(A\setminus B)$ and therefore we just have to prove $\supseteq$.

Suppose that $p\in f(A\setminus B)$ therefore $p=f\left(x\right)$ where $x\in A\setminus B$, since $x\in A$ then we know then $f\left(x\right)\in A$ but just because $x\notin B$ it doesn't directly imply that $f\left(x\right)\notin f\left(B\right)$, but if we suppose for the sake of contradiction that $p\in f\left(B\right)$ then we would deduce that $p=f\left(y\right)$ for some $y\in B$ but then we have $f\left(x\right)=p=f\left(y\right)$ and since $f$ is injective, then $x=y$ which is a contradiction, because that would imply that $x\in B$, therefore it must be the case that $p\notin f\left(B\right)$ so that $p\in f\left(A\right)\setminus f\left(B\right)$ as needed.

Let $x\in f^{-1}\left(A\right)$ so that $f\left(x\right)\in A$ we then know that $f\left(x\right)\in B$ so therefore $x\in f^{-1}\left(B\right)$

$x\in f^{-1}(\mathrm{ran}\left(f\right))$ iff $f\left(x\right)\in \mathrm{ran}\left(f\right)$ iff $x\in \mathrm{dom}\left(f\right)$ by definition

A point $p$ is an element of ${(g\circ f)}^{-1}\left(S\right)$ iff $(g\circ f)\left(p\right)\in S$ by definition this means that $g\left(f\left(p\right)\right)\in S$ iff $f\left(p\right)\in g^{-1}\left(S\right)$ iff $p\in f^{-1}\left(g^{-1}\left(S\right)\right)$ as needed.

We can observe that $p\in f^{-1}(\bigcup ℳ)$, if and only if $f\left(p\right)\in \bigcup ℳ$, iff there exists some $M\in ℳ$ such that $f\left(p\right)\in M$ iff $p\in f^{-1}\left(M\right)$ so that $p\in \bigcup \{f^{-1}\left(M\right):M\in ℳ\}$ as needed.

Since all the logical connectives in the above proof are iff's then this shows the two sets equal

$x\in f^{-1}(A\setminus B)$ iff $f\left(x\right)\in A\setminus B$ iff $f\left(x\right)\in A$ and $f\left(x\right)\notin B$, iff $x\in f^{-1}\left(A\right)$ and $x\notin f^{-1}\left(B\right)$, iff $x\in f^{-1}\left(A\right)\setminus f^{-1}\left(B\right)$, as needed.

We can observe that $p\in f^{-1}(\bigcap ℳ)$, if and only if $f\left(p\right)\in \bigcap ℳ$, iff for every $M\in ℳ$ we have $f\left(p\right)\in M$ iff $p\in f^{-1}\left(M\right)$ (still for each $M\in ℳ$ ) so that $p\in \bigcap \{f^{-1}\left(M\right):M\in ℳ\}$ as needed.

Since all the logical connectives in the above proof are iff's then this shows the two sets equal

Let $p\in f\left(f^{-1}\left(S\right)\right)$, therefore $p=f\left(x\right)$ where $x\in f^{-1}\left(S\right)$, therefore $f\left(x\right)\in S$ as needed.

Let $s\in S$, we wish to prove that $s\in f^{-1}\left(f\left(S\right)\right)$, which means that we must show $f\left(s\right)\in f\left(S\right)$, clearly this is true since $s\in S$

We've already proven that $f^{-1}\left(f\left(S\right)\right)\supseteq S$, so we just need to prove that $f^{-1}\left(f\left(S\right)\right)\subseteq S$.

Suppose that $p\in f^{-1}\left(f\left(S\right)\right)$, therefore $f\left(p\right)\in f\left(S\right)$, which means that there is some $s\in S$ such that $f\left(p\right)=f\left(s\right)$, since we assumed that $f$ was injective this implies that $p=s$, which concludes by showing $p\in S$.

We're already aware that $f\left(f^{-1}\left(S\right)\right)\subseteq S$ , so we must just show that $f\left(f^{-1}\left(S\right)\right)\supseteq S$.

So we start by assuming $s\in S$, and now we must show that $s\in f\left(f^{-1}\left(S\right)\right)$, this is only true if $s=f\left(x\right)$ for some $x\in \left(f^{-1}\left(S\right)\right)$, which in turn means that $x$ also has to satisfy: $f\left(x\right)\in S$.

We start with $s\in S\subseteq Y$, thus since $s\in Y$ and $f$ is surjective we get some $a\in X$ such that $f\left(a\right)=s$, since $s\in S$ , then also $a\in f\left(S\right)$ as needed.

We'll show that $g\circ f$ is surjective first, so let $c\in C$, since $g$ is bijective we get some $b\in B$ such that $g\left(b\right)=c$, since $b\in B$, then since $f$ is surjective we get some $a\in A$ such that $f\left(a\right)=b$, thus $g\circ f\left(a\right)=g\left(f\left(a\right)\right)=g\left(b\right)=c$ so then $g\circ f$ is surjective

Now let's show that $g\circ f$ is injective, so suppose that $a_1,a_2\in A$ and assume that $g\circ f\left(a_1\right)=g\circ f\left(a_2\right)$, note that this means $g\left(f\left(a_1\right)\right)=g\left(f\left(a_2\right)\right)$, therefore since $g$ is injective we know that $f\left(a_1\right)=f\left(a_2\right)$ and since $f$ is injective then we know that $a_1=a_2$ as needed.

Let $b_1,b_2\in B$ such that $b_1{\le }_B{b}_2$ now we want to prove that $f^{-1}\left(b_1\right){\le }_A{f}^{-1}\left(b_2\right)$.

If it so happens that $b_1=b_2$ then we automatically have that $f^{-1}\left(b_1\right)=f^{-1}\left(b_2\right)$ as ${\le }_A$ is reflexive.

So suppose rather that $b_1{\le }_B{b}_2$ and that $b_1\ne b_2$. Note that $f^{-1}\left(b_1\right)=a_1$ and $f^{-1}\left(b_2\right)=a_2$ for some $a_1,a_2\in A$ since ${\le }_A$ is a total order than $a_1,a_2$ are comparable, if it so happens that $a_2{\le }_A{a}_1$ then since $f$ is order preserving, then we would have $f\left(a_2\right){\le }_{B}f\left(a_1\right)⟺b_2{\le }_B{b}_1$ since $f$ was assumed invertible, which would only be possible if $b_1=b_2$ which we know is not the case, therefore we must have that $a_1{\le }_A{a}_2$ which means that $f^{-1}\left(b_1\right){\le }_A{f}^{-1}\left(b_2\right)$