[proofplan] We restrict the homeomorphism $f$ to the subset $A$ and prove directly that this restriction is a homeomorphism from $A$ onto $f(A)$ with their subspace topologies. Then any separation of $f(A)$ pulls back to a separation of $A$, so connectedness of $A$ forces connectedness of $f(A)$. The converse follows by applying the same argument to the inverse homeomorphism $f^{-1}: Y \to X$ and the subset $f(A) \subset Y$. [/proofplan] [step:Restrict the homeomorphism to the chosen subspace] Define the restricted map \begin{align*} g: A \to f(A), \qquad x \mapsto f(x). \end{align*} Since $f: X \to Y$ is injective, $g$ is injective. Since every element of $f(A)$ has the form $f(a)$ for some $a \in A$, $g$ is surjective. Hence $g$ is a bijection. We verify continuity of $g$ with respect to the subspace topologies. Let $W \subset f(A)$ be open in the [subspace topology](/page/Subspace%20Topology) of $f(A)$. By definition of the subspace topology, there exists $O \in \tau_Y$ such that \begin{align*} W = f(A) \cap O. \end{align*} Then \begin{align*} g^{-1}(W) = A \cap f^{-1}(O). \end{align*} Because $f$ is continuous and $O \in \tau_Y$, we have $f^{-1}(O) \in \tau_X$. Therefore $A \cap f^{-1}(O)$ is open in the subspace topology of $A$, so $g$ is continuous. The inverse map of $g$ is \begin{align*} g^{-1}: f(A) \to A, \qquad y \mapsto f^{-1}(y). \end{align*} Let $E \subset A$ be open in the subspace topology of $A$. Then there exists $U \in \tau_X$ such that \begin{align*} E = A \cap U. \end{align*} Since $f^{-1}: Y \to X$ is continuous, equivalently since $f$ is a homeomorphism, the set $(f^{-1})^{-1}(U)$ is open in $Y$. Also, \begin{align*} (g^{-1})^{-1}(E) = f(A) \cap (f^{-1})^{-1}(U). \end{align*} Thus $(g^{-1})^{-1}(E)$ is open in the subspace topology of $f(A)$, so $g^{-1}$ is continuous. Hence $g: A \to f(A)$ is a homeomorphism. [/step] [step:Pull back a separation of $f(A)$ to a separation of $A$] Assume that $A$ is connected in the subspace topology. Suppose, toward a contradiction, that $f(A)$ is not connected in the subspace topology. Then there exist subsets $U, V \subset f(A)$ such that $U$ and $V$ are open in the subspace topology of $f(A)$, both are nonempty, and \begin{align*} U \cap V = \varnothing. \end{align*} Their union is all of $f(A)$: \begin{align*} U \cup V = f(A). \end{align*} Since $g: A \to f(A)$ is continuous, the sets $g^{-1}(U)$ and $g^{-1}(V)$ are open in the subspace topology of $A$. Since $g$ is surjective and $U,V$ are nonempty, both $g^{-1}(U)$ and $g^{-1}(V)$ are nonempty. Since $U \cap V = \varnothing$, \begin{align*} g^{-1}(U) \cap g^{-1}(V) = g^{-1}(U \cap V) = g^{-1}(\varnothing) = \varnothing. \end{align*} Since $U \cup V = f(A)$, \begin{align*} g^{-1}(U) \cup g^{-1}(V) = g^{-1}(U \cup V) = g^{-1}(f(A)) = A. \end{align*} Thus $g^{-1}(U)$ and $g^{-1}(V)$ form a separation of $A$, contradicting the connectedness of $A$. Therefore $f(A)$ is connected. [guided] Assume that $A$ is connected in its subspace topology. To prove that $f(A)$ is connected, we use the contrapositive form of the definition of connectedness: if $f(A)$ were disconnected, then it would admit a separation. Thus suppose, for contradiction, that $f(A)$ is not connected in its subspace topology. Then there are subsets $U, V \subset f(A)$ such that $U$ and $V$ are open in the subspace topology of $f(A)$, neither set is empty, the two sets are disjoint, and their union is all of $f(A)$: \begin{align*} U \cap V = \varnothing. \end{align*} \begin{align*} U \cup V = f(A). \end{align*} The useful move is to transport this alleged separation back along the restricted homeomorphism \begin{align*} g: A \to f(A), \qquad x \mapsto f(x). \end{align*} Because $g$ is continuous, the preimages $g^{-1}(U)$ and $g^{-1}(V)$ are open in the subspace topology of $A$. Because $g$ is surjective onto $f(A)$, every point of $U$ and every point of $V$ has a preimage in $A$; since $U$ and $V$ are nonempty, this implies that $g^{-1}(U)$ and $g^{-1}(V)$ are nonempty. Now preimages preserve intersections and unions. The disjointness of $U$ and $V$ gives \begin{align*} g^{-1}(U) \cap g^{-1}(V) = g^{-1}(U \cap V) = g^{-1}(\varnothing) = \varnothing. \end{align*} The fact that $U$ and $V$ cover $f(A)$ gives \begin{align*} g^{-1}(U) \cup g^{-1}(V) = g^{-1}(U \cup V) = g^{-1}(f(A)) = A. \end{align*} The last equality uses that $g$ has codomain $f(A)$ and is defined on all of $A$. Therefore $g^{-1}(U)$ and $g^{-1}(V)$ are two nonempty disjoint open subsets of $A$ whose union is $A$. This is a separation of $A$, contradicting the hypothesis that $A$ is connected. Hence no separation of $f(A)$ exists, so $f(A)$ is connected. [/guided] [/step] [step:Apply the same argument to the inverse homeomorphism] Assume that $f(A)$ is connected in the subspace topology inherited from $Y$. Since $f: X \to Y$ is a homeomorphism, its inverse \begin{align*} f^{-1}: Y \to X \end{align*} is also a homeomorphism. Apply the implication already proved to the homeomorphism $f^{-1}: Y \to X$ and to the subset $f(A) \subset Y$. The image of $f(A)$ under $f^{-1}$ is \begin{align*} f^{-1}(f(A)) = A, \end{align*} where the equality follows from injectivity and surjectivity of $f$ onto its image. Therefore $A$ is connected in the subspace topology inherited from $X$. Combining the two implications, $A$ is connected if and only if $f(A)$ is connected. This proves the theorem. [/step]