# Homotopy equivalence of smash products

Suppose that $f:A\rightarrow B$ is a homotopy equivalence (both $A$ and $B$ are CW complexes), and $Y$ is a CW complex. Then is it true that the induced map

$f\wedge Id:A\wedge Y\rightarrow B\wedge Y$

is also a homotopy equivalence?

More generally, I want to answer the question: if $f$ is an $n$ connected map, that is, it induces isomorphisms $\pi_i(A)\cong\pi_i(B)$ for all $i<n$, then is the map $f\wedge Id$ also $n$ connected? i.e. is $\pi_i(A\wedge Y)\cong\pi_i(B\wedge Y)$ for all $i<n$?

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## 1 Answer

The answer to the first question is: Yes. I assume that you take the Kelley topology on the smash product, so that it becomes associative (perhaps this doesn't matter for CW complexes?). I also assume that you work with pointed CW complexes (otherwise I don't know what the smash product should be).

If $f,g : A \to B$ are homotopic, this can be realized by a homotopy $A \wedge I \to B$. This induces a map $(Y \wedge A) \wedge I \cong Y \wedge (A \wedge I) \to Y \wedge B$, which is obviously a homotopy between $Y \wedge f$ and $Y \wedge g$. It follows that $Y \wedge -$ induces an endofunctor of the homotopy category. In particular it preserves isomorphisms, i.e. homotopy equivalences.

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thanks for you answer, that makes sense. Do you know if you can say anything similar for the case when $f$ is only $n$ connected? – Moss Mar 6 '13 at 0:42
Just to be precise, a homotopy is a map $A \wedge I^{+} \rightarrow B$, where $(-)^{+}$ is a disjoint sum with a point. – Piotr Pstrągowski Mar 6 '13 at 8:39