Continuity of Composite with Identification Mapping

Theorem

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.

Let $f_1: T_1 \to T_2$ be a mapping.

Let $S_3$ be a set.

Let $p: S_1 \to S_3$ be a mapping.

Let $\tau_3$ be the identification topology on $S_3$ with respect to $p$ and $T_1$.

Let $T_3 = \struct {S_3, \tau_3}$ be the resulting topological space.

Let $f_2: S_3 \to S_2$ be a mapping such that:

$f_1 = f_2 \circ p$

where $f_2 \circ p$ is the composition of $f_2$ with $p$.

Then $f_1$ is a continuous mapping if and only if $f_2$ is a continuous mapping.

$\begin{xy} \xymatrix@L+2mu@+1em{

S_1 \ar[r]^*{p}
    \ar@{-->}[rd]_*{f_2 \circ p}

&

S_3 \ar[d]^*{f_2}

\\ &

S_2

}\end{xy}$

Proof

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Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.8$: Quotient spaces: Proposition $3.8.2$