Factoring a continuous map through the projection of an equivalence relation induced by that map yields an injective continuous map

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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

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

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces and let $f:X\rightarrow Y$ be a continuous map. Then factoring $f$ through the canonical projection of the equivalence relation induced by the mapping $f$ can not only be done, but in addition the map it yields, $\bar{f}:\frac{X}{\sim}\rightarrow Y$, is a continuous injection.

Furthermore, if $f:X\rightarrow Y$ is surjective then so is $\bar{f}:\frac{X}{\sim}\rightarrow Y$ also, making $\bar{f}$ a bijection

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 $f$ through $\pi:X\rightarrow\frac{X}{\sim}$ to get a unique (as the canonical projection of the equivalence relation is surjective) map:

• $\bar{f}:\frac{X}{\sim}\rightarrow Y$

Which is injective.

• We must show $\bar{f}$ is continuous

This is easy, simply:

[Expand]

Recall the characteristic property of the quotient topology:

$\xymatrix{ X \ar[d]_{q} \ar[dr]^{f\circ q} & \\ Y \ar[r]_f & Z }$
In this commutative diagram $f$ is continuous $\iff$ $f\circ q$ is continuous

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces and let $q:X\rightarrow Y$ be a quotient map. Then:

• For any topological space, $(Z,\mathcal{ H })$ a map, $f:Y\rightarrow Z$ is continuous if and only if the composite map, $f\circ q$, is continuous

Proof body

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The message provided is:

Would be good to show, the key thing is:

• By the characteristic property of the quotient topology we see that:
  • $f:X\rightarrow Y$ is continuous if and only if $\bar{f}:\frac{X}{\sim}\rightarrow Y$ is continuous.

So automatically, $\bar{f}$ is continuous! It's that easy!

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