What is the induced map on group completions defined by coherence sequences?

Solution 1:

You shouldn't think of coherent sequences as sequences that you want to apply maps to pointwise. Instead, you should think of a sequence $(\xi_i)$ as representing an element $\xi\in\hat{G}$ by telling you what $\xi$ is mod $G_i$ for each $i$. An element of $\hat{G}$ is like an element of $G$, but instead of having an actual element of $G$, we only know its values mod $G_i$ for each $i$ (and these values are compatible with each other).

So, if we want to define $\hat{f}(\xi)$, what do we do? Well, we need to know what $\hat{f}(\xi)$ is mod $H_i$ for each $i$, since that's what it takes to give an element of $\hat{H}$. Now by continuity of $f$, there is some $j$ such that $f(G_j)\subseteq H_i$. This means that if we know the value of $\xi$ mod $G_j$, that should be enough to know "$f(\xi)$" mod $H_i$, since our error in the input is in $G_j$ so the error in the output will be in $H_i$. So to define the value of $\hat{f}(\xi)$ mod $H_i$, we should just apply $f$ to the value of $\xi$ mod $G_j$.

That is, the $i$th term of the sequence $\hat{f}(\xi)$ is defined as follows. Let $j$ be such that $f(G_j)\subseteq H_i$, and let the $j$th term of $\xi$ be $g_j+G_j$. Then the $i$th term of $\hat{f}(\xi)$ is $f(g_j)+H_i$. (This does not depend on the choice of representative $g_j$ since $f(G_j)\subseteq H_i$, and does not depend on the choice of $j$ with this property since $\xi$ is coherent.) I will leave it to you to check that this definition actually defines a coherent sequence $\hat{f}(\xi)$ and makes $\hat{f}$ a continuous homomorphism.