# Homotopy equivalence in the category of arrows.

I'm reading Jeff Strom's book on Homotopy Theory and I am trying to make some sense of a certain exercise. On page 91, "Homotopy in Mapping Categories" we consider the category of arrows of topological spaces, that is, the objects are continuous maps $f:X \rightarrow Y$ , X and Y are topological spaces. Morphisms between two maps $f:X \rightarrow Y$, $g: Z \rightarrow W$ are pairs $\alpha = (\alpha^d,\alpha_t)$ $\alpha^d:X \rightarrow Z$, $\alpha_t : Y \rightarrow W$ making the obvious diagram commute. Let us call this category Map.

Later on, he introduces a notion of homotopy in Map. A homotopy between morphisms $\alpha:f \rightarrow g$ and $\beta:f \rightarrow g$ consists of two maps $(H^d,H_t)$, $H^d: X \times I \rightarrow Z$ , $H_t: Y \times I \rightarrow W$ , such that we have $H_t \circ (f \times id_I) = g \circ H^d$. Now, $\alpha:f \rightarrow g$ is a homotopy equivalence in Map if we can find $\beta:g \rightarrow f$ such that both $\alpha \circ \beta \simeq id_{g}$ and $\beta \circ \alpha \simeq id_{f}$. OK, so later on he explains that this is quite a strong requirement of homotopy, and one sometimes weakens it. One is pointwise homotopy equivalence in Map, which is the requirement that both $\alpha^d$ and $\alpha_t$ as above are homotopy equivalences of topological spaces. He then asks the following:

Show that if $f \simeq g:X \rightarrow Y$, then f and g are equivalent in $\textbf{h}\mathcal{T}$.

Here $\textbf{h} \mathcal{T}$ is a homotopy category of topological spaces. I am very confused by the above problem. My question is:

What does the problem ask me to do?

Does $f \simeq g$ above mean that f and g are homotopic? Or simply that they are an equivalence in Map? I am not sure what to show to be honest, and I have been trying to understand the notation for quite some time now here, so I would be beyond grateful for any help on this matter.

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Presumably, $f\cong g$ means that $f$ and $g$ are isomorphism as objects in Map. –  Rasmus Jul 23 '12 at 17:25
Oh, I see that I used $\cong$ instead of $\simeq$. I'll change that. –  Dedalus Jul 23 '12 at 17:28
Justin: Isn't that completely tautological by the definition of the homotopy category of spaces? (In some sense I guess all things in math are tautological, but still ) We just identify the homotopic maps? Thanks! –  Dedalus Jul 23 '12 at 17:36
I haven't read Strom, but these days, the homotopy category means we formally invert the weak equivalences. There is still something to prove to show that homotopic maps induce the same map, it's fairly trivial, but not tautological. –  Justin Young Jul 23 '12 at 17:39
I think the problem is asking you to show that if two particular objects of $Map$ are homotopy-equivalent in that category, then the underlying morphisms in $T$ are homotopic (or equivalently, they are equal in $hT$). A key point is that to pass from a category to its homotopy category, you only change the morphisms; you still have the same objects as before. To clarify what @JustinYoung said, it is true that we call the homotopy category of a model category the result of inverting the weak equivalences, but the old-school definition just involves passing to homotopy classes of maps. –  Aaron Mazel-Gee Jul 25 '12 at 14:55