Topology Lecture 12: Product Spaces

We derive the product topology based on the universal property we want it to fulfill. We then see how to form the product of continuous maps and list the most important properties of product spaces. In particular, we show that the product of topological manifolds is again a topological manifold.
00:00 Introduction
00:17 Derivation of the Product Topology
20:18 Definition: Product Topology
24:17 Characteristic Property of Product Topology
34:45 Definition: Product of Continuous Maps
38:04 Prop: Product of continuous maps is continuous
42:12 Prop: Product of homeomorphisms is homeomorphism
48:31 Properties of Product Spaces
52:09 Cor: Product of topological manifolds is a topological manifold
This lecture follows Lee's "Introduction to topological manifolds", chapter 3.
A playlist with all the videos in this series can be found here:
• Topology

Пікірлер: 8

  • @nathanieljones7268
    @nathanieljones72683 жыл бұрын

    Great video again! Hopefully people will start finding and appreciating your channel soon.

  • @Maria-yx4se
    @Maria-yx4se2 ай бұрын

    I started learning this subject because I thought it was about donuts, turns out its just set theory on steroids, much like abstract algebra. Now I understand the importance of set theory as a foundation for these fields

  • @shishirsanjeev4051
    @shishirsanjeev40512 жыл бұрын

    At 32:20, it is not clear to me how the union of the function is written as the union of intersection of component maps. Is it inherently assumed that f can be broken down into its component maps?

  • @mariusfurter

    @mariusfurter

    Жыл бұрын

    Sorry for the late response. I was on a boat without internet. Regarding your question, the statement is always true for the component maps. Looking at one element on the union, the claim can be reduced to f^{-1}(V_1 x ... x V_n) = f_1^{-1}(V_1) intersect ... intersect f_n^{-1}(V_n). Suppose x \in f^{-1}(V_1 x ... x V_n), then f(x) \in V_1 x ... x V_n. By definition of the component map f_i we have \pi_i(f(x)) = f_i(x). Since \pi_i maps V_1 x ... x V_n to V_i, we see that \pi_i(f(x)) = f_i(x) must lie in V_i. Hence x \in f_i^{-1}(V_i). Since i was arbitrary x lies in the intersection of the f_i^{-1}(V_i). Conversely, suppose x \in f_1^{-1}(V_1) intersect ... intersect f_n^{-1}(V_n). Then f_i(x) \in V_i for all i. Since f(x) = (f_1(x), ..., f_n(x)) this shows that x \in V_1 x ... x V_n.

  • @nasrinabas1319
    @nasrinabas1319 Жыл бұрын

    ❤❤❤

  • @Hermis14
    @Hermis142 жыл бұрын

    Can I summarize the notions of induced topology as follows?: A topology of a space constructed from existing spaces is determined by which canonical maps it makes continuous, and a characteristic property is designed to make the topology unique.

  • @mariusfurter

    @mariusfurter

    2 жыл бұрын

    I would add that the universal property allows us to use the construction without having to know exactly how it is realized. It a sort of high-level way of interacting with the objects in question, like working in a nice programming language, rather than with machine code. In fact, the maps into (or out of) a given object already completely determine it up to unique isomorphism. This is the content of the Yoneda lemma in category theory. This allows us to define objects by saying what the maps out of it (or into it) are. You probably already know many constructions that work in this way.

  • @Hermis14

    @Hermis14

    2 жыл бұрын

    ​@@mariusfurter Thank you so much for the elaborated reply. I think I am almost there but I need to enter into details of categories to understand what is really meant by the universal property.