$(X \wedge K)/(A \wedge K)= (X/A) \wedge K$

Question

I have to prove that the following equality $$(X \wedge K)/(A \wedge K)= (X/A) \wedge K$$ holds for a CW-pair $(X, A)$ and a fixed CW-complex $K$; here $\wedge$ denotes the smash product, defined as $A \wedge B=A \times B / (A \vee B)$. I am trying to see both spaces as quotients of $X \times K$, and I would like to say that I'm collapsing the same subcomplex on both sides of the equality; it should look like $A \times K \ \cup$ "something else", but I can't find a "rigorous" way to prove it.

Answer 1

There is a map $\pi : X\to X/A$ from which it follows that there is a map $\pi\wedge K: X\wedge K \to (X/A)\wedge K$. Can you show this map factors through the quotient $\pi' : X\wedge K\to (X\wedge K)/(A\wedge K)$?