# 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

 Commutative diagram showing the situation [ilmath]\xymatrix{ X \ar[r]^f \ar[d]_{\pi} & Y \\ \frac{X}{\sim} \ar@{.>}[ur]_{\tilde{f} } }[/ilmath]
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:

 In this commutative diagram[ilmath]f[/ilmath] is continuous[ilmath]\iff[/ilmath][ilmath]f\circ q[/ilmath] is continuous [ilmath]\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }[/ilmath]
Let [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath] be topological spaces and let [ilmath]q:X\rightarrow Y[/ilmath] be a quotient map. Then[2]:
• 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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So automatically, [ilmath]\bar{f} [/ilmath] is continuous! It's that easy!

We know it's injective from the factoring part mentioned above.

## Notes

1. Recall, for [ilmath]x,y\in X[/ilmath] we defined:
• [ilmath]x\sim y\iff f(x)=f(y)[/ilmath]
See: equivalence relation induced by a function