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Two maps $f,g$ into $Y$ are n-homotopic if, for every complex $K$ of dimension at most $n$ and for every map $\phi$ of $K$ into $X$ the compositions $f \phi,g\phi:K \to Y$ are homotopic.

As a sort of converse I am trying (it is an exercise in Mosher and Tangora, and hence I am confident they exist!) to find example of spaces $X$ and $Y$ and maps $f,g:X \to Y$ such that $f \phi \sim g \phi$ for any map $\phi$ of any complex $K$ into $X$ with with $f,g$ not homotopic.

Now I can never seem to develop counter-examples, and I guess I would like to understand the thought process that one goes through when developing these. So here my thought process is that $\phi$ can basically be any complex and any map, so I best concentrate my efforts on $X$ and $Y$. Initially I started thinking about the usual 'warning space' one sees with the Whitehead theorem (e.g $S^2 \times \mathbb{R}P^3$ and $\mathbb{R}P^2 \times S^3$). But this is specifically an example where we don't have a map $f:X \to Y$, so probably not relevant here. I thought some sort of maps like $S^1 \vee S^1 \to S^1$ where $f$ and $g$ collapse either one of the circles might work, but I don't think so either...

So - I'm not looking for an answer necessarily, but rather - when you see this question, what jumps into your head as a way to come up with a counter example?

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First of all $X$ should not be a complex (of dimension at most $n$), otherwise you can choose $K=X$ and $\phi = id$, showing $f \sim g$. –  Alexander Thumm Oct 6 '11 at 6:51
    
Have you already tried some related counterexamples, like the double comb or the closed topologist's sine? –  Alexander Thumm Oct 6 '11 at 6:56
    
@AlexanderThumm - the problem I have with the usual counter examples is what exactly should the map be? That's a good point about $X$ not a complex - I was hoping the example wouldn't be 'pathological' but I guess probably not! –  Juan S Oct 6 '11 at 7:09
    
The examples stated above have the property of being weakly contractible but not contractible, so what should the maps be? –  Alexander Thumm Oct 6 '11 at 7:20
    
Well weakly contractible implies homotopy groups are trivial...but I'm really not sure. So you are saying take $X$ as the double cone - I need to find two non-homotopic maps into some other space $Y$? –  Juan S Oct 6 '11 at 7:40

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