continuous images of Cauchy sequences in topological groups on page 102 of Atiyah and MacDonald's "Introduction to Commutative Algebra", they state that if $G$ and $H$ are abelian topological groups and $f$ is a continuous homomorphism from $G$ to $H$, then the image under $f$ of a Cauchy sequence in $G$ is a Cauchy sequence in $H$.
But I cannot see how this is true. This is true if we have a uniformly continuous map (between topological spaces where we have notion of uniformity), but what about here?
In fact, taking the multiplicative group of positive reals with the usual topology as $G$ and $H$, $f$ as the inverse map and $\{1/n: n=1,...\}$ as the Cauchy sequence in $G$, then its image in $H$ is the positive integers, which is not a Cauchy sequence. So the claim does not appear to be true, in general for topological groups.
What am I missing?
 A: If you have a metric on a topological group, you want uniformity: $$d(x,y)=d(ax,ay)$$ for all $a,x,y\in G$. So your metric on $\mathbb R^+$ is not uniform.
Alternatively, you can define Cauchy sequence in terms of neighborhoods of $1$ rather than $\epsilon$. A sequence $\{x_i\}$ in $G$ is Cauchy if, for any neighborhood $U$ of $1$ there exists $N$ such that if $n,m>N$ then $x_m\in x_nU$.
Under this defintion, again, your sequence is not Cauchy.
Note: The uniform metric on $(\mathbb R^+,\cdot)$ is $d(x,y)=\left|\log x - \log y\right|$. Essentially, it is the standard metric on $(\mathbb R,+)\cong (\mathbb R^+,\cdot)$.
A: A sequence $(a_n)$ in a topological group $G$ is Cauchy if it has the following property:

for every neighborhood $U$ of $1$ there exists $k$ such that, for all $m,n\ge k$, it holds $x_m x_n^{-1}\in U$.

A function $f\colon G\to H$ is uniformly continuous if, for every neighborhood $V$ of $1$ in $H$, there exists a neighborhood $U$ of $1$ in $G$ such that, whenever $ab^{-1}\in U$, we have $f(a)f(b)^{-1}\in V$.
Note: there are different notions of “uniform” for non abelian groups, where one has to distinguish between left and right uniform.
Lemma. If $f\colon G\to H$ is uniformly continuous and $(a_n)$ is a Cauchy sequence in $G$, then $(f(a_n))$ is Cauchy in $H$.
Theorem. A continuous homomorphism $f\colon G\to H$ is uniformly continuous.
Conclusion. The sequence $(1/n)$ is not Cauchy in the group of positive reals.
