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If $f$ and $g$ are homotopic maps from $X$ to $Y$ and $p : W \rightarrow X$ and $q: Y \rightarrow Z$ are any maps, then $f \circ p$ is homotopic to $g \circ p$ and $q \circ f$ is homotopic to $q \circ g$.

I see that if $H$ is homotopy from $f$ to $g$, then $H'(x,t) = q \circ H(x,t)$ is the required composition of functions since $H:X \times I\rightarrow Y$ and $q:Y \rightarrow Z$.

But I can't figure out how to find a map for $f \circ p$ and $g \circ p$. It looks likes $H(x,t) \circ p$ would be a good map, but this is impossible because $p : W \rightarrow X$ and $H:X \times I \rightarrow Y$ where the domains are not the same set. The book I'm using says $H \circ (p \times I_{d_{I}})$ is a desired map, but I am completely unaware of what $p \times I_{d_{I}}$ means for maps.

Anyone have any ideas about a map or what these symbols mean?

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The maps $p\colon W\rightarrow X$ and $q\colon Y\rightarrow Z$ are at least required to be continuous.

The map $F:=H\circ(p\times\textrm{id})$ is defined by: $$F(x,y)=H(p(x),y).$$ A proper notation could have been $H\circ(p,\textrm{id})$, but it is pretty common to write $\times$ or $\otimes$ for this operation.

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    So if $r:A \rightarrow B$ and $j:C \rightarrow D$, then $r \times j : A \times C \rightarrow B \times D$? – 2017-02-02
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    Exactly and furthermore $(r\times j)(a,c)=(r(a),j(c))$. – 2017-02-02
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    So how does this map satisfy $f \circ p$ homotopic to $g \circ p$? If they're homotopic there exists a family of maps $\{h'_t\}$ such that $h'_0 = f \circ p$ and $h'_1=g \circ p$. But $f \circ p$ and $g \circ p$ are still undefined because $f : X \times I \rightarrow Y$ and $p: W \rightarrow X$? – 2017-02-03