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If $g$ is continuous, so is the composition $g\circ p$. Conversely, assume the composition is continuous. For every open set $V\subseteq Y$,is open in $X$. The quotient definition then says that $g^{-1}(V)$ is open in $X/\mathcal{R}$. Hence $g$ is continuous andThis is the universal property of the quotient topology.

A map is continuous exactly when the preimage of every open set is open. By the definition in part (i), every open $U\subseteq X/\mathcal{R}$ has open preimage $p^{-1}(U)$. Thus $p$ is continuous, directly from the defining property of the quotient topology.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^n$ is $C^1$ on an open set and $Df(a)$ is invertible, then there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $f(a)$ such that $f:U_0\to V_0$ is a $C^1$ diffeomorphism. Its inverse satisfiesIf $f$ is $C^k$ or smooth, the inverse has the same regularity.

A topological space is compact if every open cover of it has a finite subcover. That is, if $X={\bigcup }_{i\in I}{U}_i$ with all $U_i$ open, finitely many of those same open sets already cover $X$.

For each $b\in B$, use the supplied Hausdorff condition fact to choose disjoint open sets $U_b,V_b$ with $A\subseteq U_b$ and $b\in V_b$. Compactness of $B$ gives finitely many $b_1,\dots ,b_r$ whose $V_{b_j}$ cover $B$. SetBoth are open; $U$ contains $A$ and $V$ contains $B$. They are disjoint because membership in $U\cap V$ would put a point in some $U_{b_j}\cap V_{b_j}$. If one compact set is empty, use the empty set and $X$ directly. Thus disjoint compact subspaces of a Hausdorff space admit disjoint open neighborhoods.

If $A$ is disconnected, take a separation $A=A_1\cup A_2$. Each piece is also relatively closed, being the complement of the other. They are compact as closed subsets of the compact space $A$.

Part (b), using the ambient Hausdorff property, gives disjoint ambient open neighborhoods of $A_1$ and $A_2$. They meet $A$, cover it, and are disjoint everywhere, as required. Conversely, such ambient open sets give a separation on intersecting with $A$, by part (ii). Hence compactness and the Hausdorff property permit fully disjoint ambient neighborhoods.

If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its subspace topology. By definition there are ambient open sets $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.

Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively open sets whose union is $A$. They are therefore a separation. This proves the equivalence. The ambient sets may overlap outside $A$, which is why the condition here is $U\cap V\cap A=\varnothing$.

Put $K={\bigcap }_i{F}_i$. If some $F_i$ is empty, so is $K$, which is connected. Otherwise the compact sets are closed in the Hausdorff ambient space, and can all be considered as nested nonempty closed subsets of the compact space $F_0$. Part (a)(ii) gives $K\ne \varnothing$, and $K$ is compact.

Suppose $K$ were disconnected. Part (c)(iii) gives disjoint ambient open sets $U,V$ covering $K$, both meeting it. Part (a)(iii), applied within $F_0$ to $U\cup V$, gives $F_N\subseteq U\cup V$ for some $N$. Both $U$ and $V$ meet $F_N$ because they meet $K\subseteq F_N$. Their intersections with $F_N$ would therefore disconnect $F_N$, a contradiction. Thus

A topological space is compact if every open cover of it has a finite subcover. That is, if $X={\bigcup }_{i\in I}{U}_i$ with all $U_i$ open, finitely many of those same open sets already cover $X$.

For each $b\in B$, use the supplied Hausdorff condition fact to choose disjoint open sets $U_b,V_b$ with $A\subseteq U_b$ and $b\in V_b$. Compactness of $B$ gives finitely many $b_1,\dots ,b_r$ whose $V_{b_j}$ cover $B$. SetBoth are open; $U$ contains $A$ and $V$ contains $B$. They are disjoint because membership in $U\cap V$ would put a point in some $U_{b_j}\cap V_{b_j}$. If one compact set is empty, use the empty set and $X$ directly. Thus disjoint compact subspaces of a Hausdorff space admit disjoint open neighborhoods.

If $A$ is disconnected, take a separation $A=A_1\cup A_2$. Each piece is also relatively closed, being the complement of the other. They are compact as closed subsets of the compact space $A$.

Part (b), using the ambient Hausdorff property, gives disjoint ambient open neighborhoods of $A_1$ and $A_2$. They meet $A$, cover it, and are disjoint everywhere, as required. Conversely, such ambient open sets give a separation on intersecting with $A$, by part (ii). Hence compactness and the Hausdorff property permit fully disjoint ambient neighborhoods.

If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its subspace topology. By definition there are ambient open sets $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.

Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively open sets whose union is $A$. They are therefore a separation. This proves the equivalence. The ambient sets may overlap outside $A$, which is why the condition here is $U\cap V\cap A=\varnothing$.

Put $K={\bigcap }_i{F}_i$. If some $F_i$ is empty, so is $K$, which is connected. Otherwise the compact sets are closed in the Hausdorff ambient space, and can all be considered as nested nonempty closed subsets of the compact space $F_0$. Part (a)(ii) gives $K\ne \varnothing$, and $K$ is compact.

Suppose $K$ were disconnected. Part (c)(iii) gives disjoint ambient open sets $U,V$ covering $K$, both meeting it. Part (a)(iii), applied within $F_0$ to $U\cup V$, gives $F_N\subseteq U\cup V$ for some $N$. Both $U$ and $V$ meet $F_N$ because they meet $K\subseteq F_N$. Their intersections with $F_N$ would therefore disconnect $F_N$, a contradiction. Thus

A map is continuous when the inverse image of every open set is open. Equivalently, the inverse image of every closed set is closed.

A topology $\mathcal{T}$ on $X$ contains $\varnothing$ and $X$, is closed under arbitrary unions, and is closed under finite intersections. Its members are the open sets. A subset $F\subseteq X$ is closed precisely when $X\setminus F$ is open.

A coordinate transformation is a continuously differentiable bijection from the open set $U$ onto an open subset $T(U)\subset {\mathbb{R}}^2$, with continuously differentiable inverse. Equivalently, it is a $C^1$ diffeomorphism; its Jacobian determinant never vanishes. The inverse function theorem supplies the local inverse, while global injectivity is an additional requirement.

For the canonical surjection $p:X\to X/\sim$, define a subset $W$ of $X/\sim$ to be open exactly when $p^{-1}(W)$ is open in $X$. This is the quotient topology. The empty set and the whole quotient have open inverse images. Moreover inverse images commute with arbitrary unions and finite intersections, so the corresponding axioms of a topology follow directly from those on $X$.

The map $p$ need not be an open map. For example, on $X=\mathbb{R}$ identify every point of $[0,1]$ to one point and leave all other classes singletons. The set $U=(-1/2,1/2)$ is open, butis not open. The definition of the quotient topology therefore shows that $p(U)$ is not open. The obstruction is that saturating an open set by equivalence classes need not preserve openness.

Suppose $f(X)=U\cup V$ were a separation into two disjoint nonempty relatively open sets. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$, and they are disjoint, nonempty and cover $X$. This contradicts connectedness of $X$. Hence $\boxed{f(X)\text{ is connected}}$.

The difference map $(z,w)\mapsto z-w$ is continuous and $X$ is its inverse image of the open set $\mathbb{C}\setminus \{0\}$, so $X$ is open in ${\mathbb{C}}^2$. For fixed $z$, the equivalence class is $\{z\}\times (\mathbb{C}\setminus \{z\})$, homeomorphic to a punctured plane. It is path-connected: two points can be joined by a polygonal path avoiding the missing point. Thus all equivalence classes are connected.

The projection $\pi$ is continuous and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in ${\mathbb{C}}^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi (U)$. Hence $\pi (U)$ is open.

Its fibers are precisely the equivalence classes, so it induces a bijection $\bar{\pi }:X/R\to \mathbb{C}$. Continuity follows from the quotient topology and $\pi =\bar{\pi }\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar{\pi }(V)=\pi (q^{-1}(V))$ is open. Thus the inverse is continuous too, proving $\boxed{X/R\cong \mathbb{C}}$ as topological spaces. The plane is connected, so the connected-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.

A subset $A$ is connected if, with its subspace topology, it cannot be written as $A=U\cup V$ with $U,V$ disjoint nonempty relatively open sets. Equivalently its only subsets both open and closed relative to $A$ are $\varnothing ,A$. This definition includes the empty set and singletons as connected sets.