Induced map?

1. Feb 18, 2013

dumbQuestion

I have what I hope to be just a simple notation/definition question I can't seem to find an answer to.

I'm not going to post my hw question, just a piece of it so I can figure out what the question is actually asking. I have a function i:A --> X I also have a continuous function g: A --> B. Then I am asked to prove a property about the "induced map" f: B --> B Ug X

I am just having trouble understanding exactly what this "induced map" is. There are no defs for it in my book and online I only see induced map as induced homeomorphisms. So my question: is this a quotient map? Is B Ug X just a quotient space? My map i is not necessarily an inclusion, so A and X could be two separate spaces, so I"m assuming this is a quotient space because an a in A could map to X under i but could also map to B under g. I guess I'm just confused about this function. Does i even factor in to this map?

Also when I think of the composition f(i(A)) what on earth this would be like.

2. Feb 18, 2013

lavinia

Take the disjoint union of B and X modulo the equivalence relation g(a)= i(a).

The f is just the identity map on B.

This seems misstated? are you sure this is right?

3. Feb 18, 2013

lavinia

the only thing I could see is that i allows you to form the equivalence relation. When you identify i(a) and g(a) you glue B to X. This glued together space probably has the quotient topology.

What is a cofibration?

4. Feb 18, 2013

dumbQuestion

I think you're right about what the map is doing, that's what I was thinking myself.

cofibration is another mess... its just a map that has the homotopy extension property for every space Y. So, i : A --> X is a cofibration if, for every homotopy H_t : A --> Y and every map M: X --> Y that agrees with H_0 on A, then you can extend the homotopy H_t to one that goes from X --> Y and agrees on A.

5. Feb 18, 2013

lavinia

then i think the construction is as we think.