Existence and Uniqueness of Direct Limit of Sequence of Groups/Lemma 3

From ProofWiki
Jump to navigation Jump to search


Let $h_\infty: G_\infty \to H$ be the mapping defined as:

$\eqclass {\tuple {x_n, n} } {} \mapsto \map {h_n} {x_n}$

Then $h_\infty$ is a well-defined group homomorphism.


Well-Definedness of $h_\infty$

Let $\tuple {x_n, n}, \tuple{x_{n'}, n'} \in \eqclass {\tuple {x_n, n} } {}$.

Without loss of generality, let $n' \ge n$.

Then we have:

$\map {g_{n, n'} } {x_n} = x_{n'}$


\(\ds \map {h_{n'} } {x_{n'} }\) \(=\) \(\ds \map {h_{n'} } {\map {g_{n,n'} } {x_n} }\)
\(\ds \) \(=\) \(\ds \map {\paren {h_{n'} \circ g_{n, n'} } } {x_n}\) because $h_{n'} \circ g_{n, n'} = h_n$
\(\ds \) \(=\) \(\ds \map {h_n} {x_n}\)

This proves that $h_\infty$ is independent of the representative chosen.

That is, $h_\infty$ is well-defined.


Homomorphism Property

Let $\eqclass {\tuple{x_n, n} } {}, \eqclass {\tuple {y_m, m} } {} \in G_\infty$.

Without loss of generality, we may assume, by the definition of the group operation, that $n = m$.

See Lemma 2 for details.

It follows that:

\(\ds \map {h_{\infty} } {\eqclass {\tuple {x_n, n} } {} \circ \eqclass {\tuple {y_n, n} } {} }\) \(=\) \(\ds \map {h_n} {x_n y_n}\) Definition of $h_\infty$
\(\ds \) \(=\) \(\ds \map {h_n} {x_n} \map {h_n} {y_n}\) $h_n$ is a homomorphism
\(\ds \) \(=\) \(\ds \map {h_\infty} {\eqclass {\tuple {x_n, n} } {} } \circ \map {h_\infty} {\eqclass {\tuple {y_n, n} } {} }\) Definition of $h_\infty$

Thus $h_\infty$ is a homomorphism.