2022-11-13

393: Restriction of Continuous Map on Domain and Codomain Is Continuous

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that restriction of continuous map on domain and codomain is continuous

Topics


About: topological space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.


Main Body


1: Description


For any topological spaces, T1,T2, any continuous map, f:T1T2, any subset, S1T1, and any subset, S2T2 such that f(S1)S2, f|S1:S1S2 is continuous.


2: Proof


For any open set, US2, U=US2 where UT2 is open on T2, by the definition of subspace topology. f|S11(U)=f|S11(US2)=f1(U)S1, because for any pf|S11(US2), f|S1(p)US2 by the definition of preimage, f(p)US2U, so, pf1(U), while of course pS1 because S1 is the domain of f|S11; for any pf1(U)S1, f|S1(p)U because as p is on S1, f|S1 can operate on it with the same result, which is on U, but as f|S1(S1)=f(S1)S2, f|S1(p)US2, so, pf|S11(US2). As f is continuous, f1(U) is open on T1, and f1(U)S1 is open on S1 by the definition of subspace topology.


References


<The previous article in this series | The table of contents of this series | The next article in this series>