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Suppose $ f : X \to Y$ and $g,h : Y \to Z$ are continuous, with $g \simeq h$. Prove that $ gf \simeq hf $.

My attempt:

Suppose $ L(x,t) $ gives a homotopy from $g$ to $h$, i.e. $ L(x,t) : Y \times I \to Z $, with $ L(x,0) = g(x)$ and $ L(x,1) = h(x) \ \forall x \in X$.

Then let $ H(x,t) : X \times I \to Z $ be defined by $ H(x,t) = L(f(x),t) $. Then $H$ is continuous, since it is the composition of continuous maps. $H(x,0) = L(f(x),0) = g(f(x)) $ and $ H(x,1) = L(f(x),1) = h(f(x)) $. This $H$ is the desired homotopy.

Is this correct?

Thanks!

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Yes, it is. But, depending on which level you are, maybe you should write more explicitely $H$ as $H = L \circ (f\times \mathrm{id})$ in order to show that indeed it is a composition of continuous maps. –  a.r. Oct 11 '11 at 15:17
    
Is that the identity on $X$ or on $I$? I've never seen this notation before. Thanks –  TRY Oct 11 '11 at 15:23
    
On $I$. It's the map $f\times \mathrm{id} : X \times I \longrightarrow Y\times I$ defined by $(f\times \mathrm{id})(x,t) = (f(x), t)$. If you haven't seen this notation before, maybe you should prove that, whenever you have continuous maps $f: X\longrightarrow Y$ and $g: A \longrightarrow B$, then the map $f\times g : X\times A \longrightarrow Y\times B$, defined by $(f\times g)(x, a) = (f(x), g(a))$, is also continuous. –  a.r. Oct 11 '11 at 15:47
    
Great, that makes sense. I'll prove that now, and adopt this notation for the future. Thanks a lot –  TRY Oct 11 '11 at 15:54
    
I added a community wiki answer just saying this is correct (if you want to accept that, so this question does not appear as 'unanswered') –  Juan S Oct 12 '11 at 2:03
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1 Answer 1

Yes, this is correct $${}$$ ${}$

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Ideally, the accepted answer will contain the details... –  Mariano Suárez-Alvarez Oct 12 '11 at 2:07
6  
@Mariano - it is CW so feel free to edit...But the question is 'Is this correct' and the answer is indeed, yes - I, personally, am not sure there is much to add –  Juan S Oct 12 '11 at 5:35
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