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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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.
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? –  gabbering 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