[proofplan] A uniform homeomorphism transfers Cauchy sequences in both directions because both $f$ and $f^{-1}$ are uniformly continuous. If $X$ is complete, a [Cauchy sequence](/page/Cauchy%20Sequence) in $Y$ pulls back under $f^{-1}$ to a convergent sequence in $X$, and [uniform continuity](/page/Uniform%20Continuity) of $f$ sends its limit back to a limit in $Y$. The reverse implication is the same argument with $f$ and $f^{-1}$ interchanged. [/proofplan]

[step:Record the two elementary consequences of uniform continuity]We first note two sequence facts that will be used for the maps $f:X\to Y$ and $f^{-1}:Y\to X$. Let $(M,d_M)$ and $(N,d_N)$ be metric spaces, and let $g:M\to N$ be uniformly continuous. If $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence in $M$, then $(g(a_n))_{n=1}^{\infty}$ is a Cauchy sequence in $N$. Indeed, let $\varepsilon>0$. By uniform continuity of $g$, there exists $\delta>0$ such that for all $u,v\in M$, \begin{align*} d_M(u,v)<\delta \implies d_N(g(u),g(v))<\varepsilon. \end{align*} Since $(a_n)_{n=1}^{\infty}$ is Cauchy in $M$, there exists $N_0\in\mathbb N$ such that for all $m,n\ge N_0$, \begin{align*} d_M(a_m,a_n)<\delta. \end{align*} Therefore, for all $m,n\ge N_0$, \begin{align*} d_N(g(a_m),g(a_n))<\varepsilon, \end{align*} so $(g(a_n))_{n=1}^{\infty}$ is Cauchy in $N$. If $a_n\to a$ in $M$, then $g(a_n)\to g(a)$ in $N$. Indeed, let $\varepsilon>0$ and choose $\delta>0$ from uniform continuity as above. Since $a_n\to a$, there exists $N_1\in\mathbb N$ such that for all $n\ge N_1$, \begin{align*} d_M(a_n,a)<\delta. \end{align*} Hence, for all $n\ge N_1$, \begin{align*} d_N(g(a_n),g(a))<\varepsilon, \end{align*} which proves $g(a_n)\to g(a)$.[/step]

[guided]We isolate the only mechanism used in the proof: a uniformly continuous map preserves both Cauchy sequences and convergent sequences. Let $(M,d_M)$ and $(N,d_N)$ be metric spaces, and let $g:M\to N$ be uniformly continuous. Uniform continuity means that for every tolerance $\varepsilon>0$ in the target metric $d_N$, there exists a tolerance $\delta>0$ in the source metric $d_M$ such that for every pair $u,v\in M$, \begin{align*} d_M(u,v)<\delta \implies d_N(g(u),g(v))<\varepsilon. \end{align*} First suppose $(a_n)_{n=1}^{\infty}$ is a Cauchy sequence in $M$. We must prove that $(g(a_n))_{n=1}^{\infty}$ is Cauchy in $N$. Let $\varepsilon>0$ be given. Choose $\delta>0$ by uniform continuity of $g$. Since $(a_n)_{n=1}^{\infty}$ is Cauchy in $M$, there exists $N_0\in\mathbb N$ such that whenever $m,n\ge N_0$, \begin{align*} d_M(a_m,a_n)<\delta. \end{align*} Applying the uniform continuity implication to the pair $u=a_m$ and $v=a_n$ gives, for all $m,n\ge N_0$, \begin{align*} d_N(g(a_m),g(a_n))<\varepsilon. \end{align*} This is exactly the Cauchy condition for $(g(a_n))_{n=1}^{\infty}$ in $N$. Now suppose $a_n\to a$ in $M$. We prove $g(a_n)\to g(a)$ in $N$. Let $\varepsilon>0$ be given, and choose $\delta>0$ again from uniform continuity of $g$. Since $a_n\to a$ in $M$, there exists $N_1\in\mathbb N$ such that for every $n\ge N_1$, \begin{align*} d_M(a_n,a)<\delta. \end{align*} Applying uniform continuity with $u=a_n$ and $v=a$ gives, for every $n\ge N_1$, \begin{align*} d_N(g(a_n),g(a))<\varepsilon. \end{align*} Therefore $g(a_n)\to g(a)$ in $N$. The important point is that uniform continuity is stronger than needed for convergence, but it is exactly what is needed to preserve Cauchy sequences before a limit is known.[/guided]

[step:Use completeness of $X$ to prove completeness of $Y$] Assume $(X,d_X)$ is complete. Let $(y_n)_{n=1}^{\infty}$ be an arbitrary Cauchy sequence in $Y$. Define a sequence $(x_n)_{n=1}^{\infty}$ in $X$ by \begin{align*} x_n = f^{-1}(y_n) \end{align*} for each $n\in\mathbb N$. Since $f^{-1}:Y\to X$ is uniformly continuous and $(y_n)_{n=1}^{\infty}$ is Cauchy in $Y$, the first sequence fact from the previous step implies that $(x_n)_{n=1}^{\infty}$ is Cauchy in $X$. Because $(X,d_X)$ is complete, there exists $x\in X$ such that $x_n\to x$ in $X$. Since $f:X\to Y$ is uniformly continuous, the second sequence fact from the previous step implies that $f(x_n)\to f(x)$ in $Y$. Let $\operatorname{id}_Y:Y\to Y$ denote the identity map on $Y$. For every $n\in\mathbb N$, the definition of $x_n$ and the identity $f\circ f^{-1}=\operatorname{id}_Y$ give \begin{align*} f(x_n)=f(f^{-1}(y_n))=y_n. \end{align*} Thus $y_n\to f(x)$ in $Y$. Since every Cauchy sequence in $Y$ converges in $Y$, the [metric space](/page/Metric%20Space) $(Y,d_Y)$ is complete. [/step]

[step:Use completeness of $Y$ to prove completeness of $X$] Assume $(Y,d_Y)$ is complete. Let $(x_n)_{n=1}^{\infty}$ be an arbitrary Cauchy sequence in $X$. Since $f:X\to Y$ is uniformly continuous, the first sequence fact from the first step implies that $(f(x_n))_{n=1}^{\infty}$ is a Cauchy sequence in $Y$. By completeness of $(Y,d_Y)$, there exists $y\in Y$ such that $f(x_n)\to y$ in $Y$. Define $x\in X$ by \begin{align*} x=f^{-1}(y). \end{align*} Since $f^{-1}:Y\to X$ is uniformly continuous, the second sequence fact from the first step implies that $f^{-1}(f(x_n))\to f^{-1}(y)$ in $X$. Let $\operatorname{id}_X:X\to X$ denote the identity map on $X$. For every $n\in\mathbb N$, the identity $f^{-1}\circ f=\operatorname{id}_X$ gives \begin{align*} f^{-1}(f(x_n))=x_n. \end{align*} Therefore $x_n\to x$ in $X$. Since every Cauchy sequence in $X$ converges in $X$, the metric space $(X,d_X)$ is complete. [/step]

[step:Combine the two implications] We have proved that completeness of $(X,d_X)$ implies completeness of $(Y,d_Y)$, and that completeness of $(Y,d_Y)$ implies completeness of $(X,d_X)$. Hence $(X,d_X)$ is complete if and only if $(Y,d_Y)$ is complete. [/step]