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Let $U$ be open in $Y$, we want to prove that $f^{-1}\left(U\right)$ is an open set. Since $U$ is open then we know that $U=\bigcup 𝒞$ where $𝒞\subseteq ℬ$.

By noting that $f^{-1}\left(U\right)=f^{-1}(\bigcup 𝒞)$ $=$ $\bigcup \{f^{-1}\left(C\right):C\in 𝒞\}$, since $𝒞\subseteq ℬ$, then every element $C$ is a basis element, and thus $f^{-1}\left(C\right)$ is an open set by our assumption, therefore as it is an arbitrary union of open sets, we can see that $f^{-1}\left(U\right)$ is an open set, as needed

Let $p\in A$ now let's an open set around $x$ contained in $A$. Since $p\in A$ then we know that $f\left(p\right)>g\left(p\right)$ therefore since $Y$ is hausdorff then we know that there are disjoint neiborhoods $U,V$ of $g\left(p\right),f\left(p\right)$ respectively such that $\forall u,v\in U,V$ we have $u<v$. Since $f,g$ are continuous we also knot that $S=g^{-1}\left(U\right)\cap f^{-1}\left(V\right)$ is open in $X$, if $s\in S$ then we know that $g\left(s\right)\in U$ and that $f\left(s\right)\in V$ and so $g\left(s\right)<f\left(s\right)$ therefore $s\in A$ so that $S\subseteq A$ and therefore it is open.

Suppose that $U\in {𝒯}_2$, then since ${\mathrm{id}}^{-1}=\mathrm{id}$ then we know that ${\mathrm{id}}^{-1}\left(U\right)=U$ is open in ${𝒯}_1$ since $\mathrm{id}$ is continuous, therefore we are done

Since ${𝒯}_Y$ is generated by $S$, this means that ${𝒯}_Y={𝒯}_{{ℬ}_S}$ where ${ℬ}_S$ is the basis generated by $S$ therefore to verify that this function is continuous all we have to do is check that every basis element $B\in {ℬ}_S$ has the property that $f^{-1}\left(B\right)$ is open with respect to $X$

We know $B=S_1\cap S_2\cap \dots \cap S_k$ for some $k\in {\mathbb{N}}_1$, therefore we want to know if $f^{-1}(S_1\cap S_2\cap \dots \cap S_k)$ is open, since we know that the intersection factors through the inverse image then we can re-write this as ${\bigcap }_{i=1}^k{f}^{-1}\left(S_i\right)$.

Our initial assumption was that given any $S\in 𝒮$ that $f^{-1}\left(S\right)$ was open, thus specifically for our $S_1,\dots ,S_k$ their inverse images should also be open, therefore ${\bigcap }_{i=1}^k{f}^{-1}\left(S_i\right)$ is a finite intersection of open sets of $X$ thus it is open, which shows that $f$ is continuous.

$⟹$ Let $x\in X$ and suppose $V$ is a neighborhood of $f\left(x\right)$, then $f^{-1}\left(V\right)$ is an open set since $f$ was assumed continuous, it also must contain $x$, this is because $f\left(x\right)\in V$, making $f^{-1}\left(V\right)$ a neighborhood of $x$, we also know $f\left(f^{-1}\left(V\right)\right)\subseteq V$, as needed.

$⟸$ Suppose $V$ is open in $Y$, if $f^{-1}\left(V\right)=\varnothing$, we are done, thus we can find some element $x\in f^{-1}\left(V\right)$, making $V$ a neighborhood of $f\left(x\right)$, thus by assumption, we get some neighborhood $U_x$ such that $f\left(U_x\right)\subseteq V$, this by definition means that for any $p\in U_x$, we know $f\left(p\right)\in V$, or that $p\in f^{-1}\left(V\right)$, which shows that $U_x\subseteq f^{-1}\left(V\right)$.

Since this is true for every point $x\in f^{-1}\left(V\right)$, since ${\bigcup }_{x\in f^{-1}\left(V\right)}{U}_x=f^{-1}\left(V\right)$, then $f^{-1}\left(V\right)$ can be written as an arbitrary union of open sets, and thus it is open, as needed.

Suppose that $f$ is a homeomorphism, and that $U$ is a subset of $X$, assume $f\left(U\right)$ is open in $Y$ , we'll prove that $U$ is open in $X$, since $f$ is continuous, then we know that $f^{-1}\left(f\left(U\right)\right)$ is an open set of $X$, since $f$ is a bijection, this means it's injective and thus $f^{-1}\left(f\left(U\right)\right)=U$ as needed.

Now suppose that $U$ is open in $X$, since we know that $f^{-1}$ is continuous this means that ${\left(f^{-1}\right)}^{-1}\left(U\right)$ is an open set of $Y$, we've shown that ${\left(f^{-1}\right)}^{-1}\left(U\right)=f\left(U\right)$, thus, we've just shown that $f\left(U\right)$ is an open set of $Y$, this concludes

At this point we've proven the main rightward implication. The other direction is quite a lot easier. We have to show that both $f$ and $f^{-1}$ are continuous.

Suppose that $V$ is an open set of $Y$, we want to show that $f^{-1}\left(V\right)$ is an open set of $X$, which is true iff $f\left(f^{-1}\left(V\right)\right)$ is an open set of $Y$, since $f$ is surjective, then $f\left(f^{-1}\left(V\right)\right)=V$ making it an open set of $Y$, which shows that $f^{-1}\left(V\right)$ is an open set of $X$

Now suppose that $U$ is an open set of $X$ , our goal will be to show that ${\left(f^{-1}\right)}^{-1}\left(U\right)$ is an open set of $Y$, as seen previously ${\left(f^{-1}\right)}^{-1}\left(U\right)=f\left(U\right)$, and since $U$ was open in $X$, by our assmption we know that $f\left(U\right)$ is open in $Y$, this concludes the proof.

Suppose that $V\subseteq Y$, if $p\in V$, then $f^{-1}\left(V\right)=X$, otherwise $p\notin V$ so $f^{-1}\left(V\right)=\varnothing$, in either case we have an open set of $X$.

Let $U$ be an open set of $X$, then ${\iota }^{-1}\left(U\right)=U\cap A$, which by definition is open in the subspace topology of $A$ as needed.

Let $U$ be open in $Z$, we want to prove that ${(g\circ f)}^{-1}\left(U\right)$ is an open set of $X$.

To start if we look at ${(g\circ f)}^{-1}\left(U\right)$, we know this is equal to $f^{-1}\left(g^{-1}\left(U\right)\right)$, thus since $g$ is continuous we know that $S:=g^{-1}\left(U\right)$ is an open set of $Y$ and then $f^{-1}\left(S\right)$ is open in $X$ since $f$ was continuous, this shows that $g\circ f$ is continuous

We note that $f\upharpoonright A=f\circ \iota$, therefore as the composition of two continuous functions, we know that $f\upharpoonright A$ is continuous

To show continuity we start by assuming that $B$ is open in $Z$, therefore $B=Z\cap U$ where $U$ is open in $Y$

TODO: Add the proof here.