# Homotopic maps $(X, A) \to (Y, B)$ induce homotopic maps $X/A \to X/B$

Working with a homework problem:

$X$ and $Y$ are topological spaces and $A\subset X$, $B\subset Y$. I have $f,g:(X,A)\to (Y,B)$ as homotopic maps.

I need to show that the induced maps: $\hat{f},\hat{g}$ are homotopic.

My instinct says that the homotopy I should use is $\hat{F}:I\times (X/A)\to (Y/B)$ should be given by $\hat{F}(t,[x]) := [F(t,x)]$. I showed this map is well defined without any problem.

Edit: More precisely, $\hat{f}$ maps the pair $(t,[x])$, where $x$ is any element in $[x]\in X/A$, to $[y]$, where $y$ is any element in $[F(t,x)]$.

But I'm having difficulty showing the map is continuous. If I take $V$ to be an open subset of $Y/B$, how can I use the continuity of $F$ to show that $\hat{F}$ is continuous?

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fixed. I hope it is more clear now. –  roo Feb 2 '12 at 2:29
Oh, cool. I added what I thought was enough detail but this is even better :). –  Dylan Moreland Feb 2 '12 at 2:39

Here's something you could use: the topology on $X/A$ comes from the quotient map $q\colon X \to X/A$. We know how to define continuous maps out of quotient spaces, and so it would be helpful if $\operatorname{id} \times q\colon I \times X \to I \times X/A$ were a quotient map. And indeed it is, because $I$ is locally compact: this is a theorem, the proof of which is surprisingly long, of Whitehead's. See Lemma 2.88 here.

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It's good to know that in general the product of two quotient maps need not be a quotient map. I've graded enough algebraic topology papers to know that this is an attractive fallacy. Anyway, hopefully there is an easier way of showing your result, but this is a nice result to have in your toolbelt, I think. –  Dylan Moreland Feb 2 '12 at 2:06
Your remark is actually a hint at the beginning of my assignment. :) –  roo Feb 2 '12 at 2:26