Claim: When $X ×Y$ is endowed with the product topology $T_{X×Y}$ , the projection maps $p_X : X × Y → X , p_X(x, y) = x$, and $p_Y : X × Y → Y , p_Y (x, y) = y$ , are continuous.

Proof: Indeed for any open set $U$ in $X$ , $p^{−1}_X (U) = U × Y$

This is a claim from my lecture notes. However, the projection map is clearly not injective. How can we define the preimage of a non-injective map? Clearly, $p_X(x,y_1)=x=p_X(x, y_2)$ for any $y_1\neq y_2$


For any function $f$ an any set $A$ in the range the notation $f^{-1}(A)$ stands for all points $x$ in the domain of $f$ such that $f(x)\in A$. If $f$ happens to have an inverse then $f^{-1}(A)$ becomes the image of $A$ under $f^{-1}$.

  • $\begingroup$ I see. So we are defining a map $f^{-1}:X \rightarrow X$ x $Y$ by $f^{-1}(t)=t$ x $Y$ rather than taking the inverse of $f$, is that correct? Will mark the answer as correct when the website allows me to! $\endgroup$ – asdf Dec 5 '18 at 9:42

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