Factoring a continuous map through the projection of an equivalence relation induced by that map yields an injective continuous map
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Contents
- This theorem is a minor extension of "factoring a function through the projection of an equivalence relation induced by that function yields an injection" by simply considering continuity in addition.
Statement
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be a continuous map. Then factoring [ilmath]f[/ilmath] through the canonical projection of the equivalence relation induced by the mapping [ilmath]f[/ilmath] can not only be done, but in addition the map it yields, [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath], is a continuous injection[1].Furthermore, if [ilmath]f:X\rightarrow Y[/ilmath] is surjective then so is [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] also, making [ilmath]\bar{f} [/ilmath] a bijection[Note 1]
Proof
Overview:
We know already (from factoring a function through the projection of an equivalence relation induced by that function yields an injection) that we can factor [ilmath]f[/ilmath] through [ilmath]\pi:X\rightarrow\frac{X}{\sim} [/ilmath][Note 2] to get a unique (as the canonical projection of the equivalence relation is surjective) map:
- [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath]
Which is injective.
- We must show [ilmath]\bar{f} [/ilmath] is continuous
This is easy, simply:
Recall the characteristic property of the quotient topology:
topological spaces and let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map. Then[2]:
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be - For any topological space, [ilmath](Z,\mathcal{ H })[/ilmath] a map, [ilmath]f:Y\rightarrow Z[/ilmath] is continuous if and only if the composite map, [ilmath]f\circ q[/ilmath], is continuous
Proof body
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Would be good to show, the key thing is:
- By the characteristic property of the quotient topology we see that:
- [ilmath]f:X\rightarrow Y[/ilmath] is continuous if and only if [ilmath]\bar{f}:\frac{X}{\sim}\rightarrow Y[/ilmath] is continuous.
So automatically, [ilmath]\bar{f} [/ilmath] is continuous! It's that easy!
We know it's injective from the factoring part mentioned above.Notes
- ↑ See: If a surjective continuous map is factored through the canonical projection of the equivalence relation induced by that map then the yielded map is a continuous bijection
- ↑ Recall, for [ilmath]x,y\in X[/ilmath] we defined:
- [ilmath]x\sim y\iff f(x)=f(y)[/ilmath]
References