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The problem is:

Let $X$ be a topological space, let $Y$ be a set (without a topology as yet), and suppose $f: X \rightarrow Y$ is a function (of sets).

  1. Show that $Y$ has a unique largest topology for which $f$ is continuous.
  2. Show that with this topology, a function $g: Y \rightarrow Z $ from $Y$ to some other topological space is continuous iff $g \circ f: X \rightarrow Z$ is continuous.

My question is:

1- What does this problem want to teach us and how to prove it?

TheHolyJoker
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Intuition
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    It is defining the final topology. For (a) define $U\subset Y$ to be open iff $f^{-1}(U)$ is open in $X$. Check that this collection of sets is a topology on $Y$. It is clear that you cannot add any more subsets of $Y$ to it without making $f$ discontinuous. – MoonLightSyzygy Jan 10 '20 at 05:08
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    Answers can be found in this post of mine about final topologies in general. Yours is a special case. – Henno Brandsma Jan 10 '20 at 07:07
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    what have you tried? – supinf Jan 10 '20 at 09:54
  • I do not know how to proceed at least in (a), @supinf – Intuition Jan 10 '20 at 11:42
  • @HennoBrandsma why mine is a special case? – Intuition Jan 10 '20 at 11:45
  • @MoonLightSyzygy why we defined $U$ by this way? Is not this the definition of the continuity of $f$, but $X$ and $Y$ should be topological spaces in thayt definition. why it is clear that I cannot add any more subsets of $Y$ without makinf $f$ discontinuous? – Intuition Jan 10 '20 at 11:55
  • The definition of the open sets of $Y$ look like the definition of continuity for $f$ exactly because that is goal; to build a topology such that $f$ is continuous and moreover call open as many subsets of $Y$ as possible. – MoonLightSyzygy Jan 10 '20 at 12:00
  • You cannot add any more subsets because if you add a set $V\subset Y$ such that $f^{-1}(V)$ is not open in $X$, then by definition $f$ wouldn't be continuous. – MoonLightSyzygy Jan 10 '20 at 12:01
  • Is this because yours contain $Y_{i}$ but mine contains one $Y$ only? but my $Y$ is just a set, it is not a topological space @HennoBrandsma – Intuition Jan 10 '20 at 12:37
  • @HennoBrandsma I think your post answers only part (a) .... am I correct? – Intuition Jan 10 '20 at 12:43
  • @HennoBrandsma I think the part of your answer that answers my (a) is existence theorem for final topologies ... am I correct? – Intuition Jan 10 '20 at 13:07
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    It answers both. The characterisation of continuity is also there. And because we only have one map instead of a family it’s a special case. – Henno Brandsma Jan 10 '20 at 17:28

1 Answers1

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The problem wants to show that is a natural way to endow a set $Y$ with a topology when we have a function $f: X \to Y$ where $X$ already has a topology. This obeys a nice universal property (your $2$.) and is essentially unique with that property.

There also is a unique smallest topology that makes $f$ continuous but this will always be the indiscrete topology $\{\emptyset, Y\}$ and says nothing of interest about $f$, really.

For existence: just check the axioms of topology for

$$\mathcal{T}=\{O \subseteq Y: f^{-1}[O] \text{ open in } X\}$$

using identities like $$f^{-1}[\bigcup_i O_i]=\bigcup_i f^{-1}[O_i]$$ and similar ones for intersection, plus the fact that $f^{-1}[\emptyset]=\emptyset$ and $f^{-1}[Y]=X$ for any map $f:X \to Y$ too.

This obeys 1. because if $\mathcal{T}'$ is any arbitrary topology on $Y$ that makes $f$ continuous, then for all $O \in \mathcal{T}'$ we have that $f^{-1}[O]$ is open in $X$ (by that assumed continuity!) and so by definition, $O \in \mathcal{T}$, hence $\mathcal{T}' \subseteq \mathcal{T}$ has been shown and $\mathcal{T}$ is maximal as required.

Now if $g: Y \to Z$ is continuous, then $g \circ f$ is continuous as a composition of continuous functions.

Conversely, if $g: Y \to Z$ is a function such that $g \circ f$ is continuous, let $O$ be open in $Z$. Then by the assumed continuity $$(g \circ f)^{-1}[O] = f^{-1}[g^{-1}[O]]$$ is open in $X$ and thus by definition of $\mathcal{T}$ again we have $g^{-1}[O] \in \mathcal{T}$. As $O$ was arbitrary, $g$ is continuous.

Henno Brandsma
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  • If you have time could u look at the same above question but with arrows turned around https://math.stackexchange.com/questions/3507131/show-that-f-has-a-unique-smallest-topology-for-which-f-is-continuous – Intuition Jan 13 '20 at 04:13
  • @Secretly that’s the dual statement, also in category theory sense. – Henno Brandsma Jan 13 '20 at 04:48
  • Could you please look at this question (if you have time) https://math.stackexchange.com/questions/3518158/the-sum-of-two-sets-and-the-disjoint-union ? – Emptymind Jan 23 '20 at 08:46