Quotient Map Theorem: Topology Induced by f

[ForMyThunder]

Here is theorem 9.2 from Stephen Willard's General Topology:

If X and Y are topological spaces and f:X\to Y is continuous and either open or closed, then the topology \tau on Y is the quotient topology induced by f.

So f has to be onto doesn't it? Otherwise there will be multiple topologies on Y that satisfy the hypotheses but are not the quotient topology?

 

[micromass]

Yes, f has to be onto. This must be because \tau_f is only defined for surjective maps.

And it is also used when they say f[f^{-1}(U)]=U...

 

[Landau]

I advise you to do exercise 9H. Given any map f:X\to Y, with X a top. space, we can give Y the strongest topology that makes f continuous: {U in Y whose inverse image under f is open in X}. More generally this can be done for a collection of maps f_i:X_i\to Y, and take {U in Y for which the inverse image under each f_i's is open in X}. In the terminology of Willard, say that such a family of maps {f_i}_i "covers points of Y" iff each y in Y is in the image of some f_i. In this case, the topology just described is called the "quotient topology".

In particular, if the collection {f_i}_i consists of a single function, it covers points iff it is surjective. So this reduces to the definition that if f:X\to Y is surjective, then the strongest topology making f continuous is the quotient topology.


I guess my point is: the exact same construction can be done for any map, surjective or not. But if it is surjective, then we call it the quotient topology, because this amounts to dividing out an equivalence relation, i.e. 'glueing'.

 

[ForMyThunder]

But the strongest topology on Y that makes $f:X\to Y$ continuous immediately declares that all point sets outside the image of f are open. But the theorem as stated could declare that any subset of $Y-f(X)$ is open and still be the quotient topology. Okay, I see in the definition of the quotient topology it specifies an open mapping. Thanks.

 

[Landau]

But the strongest topology on Y that makes $f:X\to Y$ continuous immediately declares that all point sets outside the image of f are open.

That's true.

But the theorem as stated could declare that any subset of $Y-f(X)$ is open and still be the quotient topology. Okay, I see in the definition of the quotient topology it specifies an open mapping.

You mean onto?

 

[ForMyThunder]

Yep, my B.