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If i have a product topology generated by two topological spaces X and Y and another product topology generated by the same sets, but with different topology(call that X' and Y'), such that XxY is finer than X'xY', is it true that also the topology of X have to be finer than that of X' and the topology of Y have to be finer than that of Y'?(I know that the converse is true)

Lorenzo Biagi
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The answer is yes. Assume, towards a contradiction, that the topology of $X$ were not finer than that of $X'$, and pick some $U$ that is open in $X'$ but not in $X$. Use that $Y$ is open in $Y$ for any topology (so in $Y$ and in $Y'$). But then $U\times Y'=U\times Y$ is open in $X'\times Y'$, but not open in $X\times Y$. Indeed, by definition of the product topology, if $U\times Y$ were open in $X\times Y$ then $U$ must be open in $X$.

(Formally you could also use that projections are open maps. See e.g. Projection is an open map
If $U\times Y$ were open in $X\times Y$ then $U$ would be open in $X$, as $U$ is the projection of the open $U\times Y$.)

Mirko
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    The last step, "$U$ is the projection of ... $U\times Y$", uses that $Y$ is nonempty. Fortunately, it is nonempty because the question says that $Y'$ has a different topology on the same set, and the empty set has only one topology. – Andreas Blass Jul 24 '19 at 02:13
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    @AndreasBlass Thank you, admittedly I didn't even think of the case when $Y$ might be empty, so this is a gap in my answer, if $Y$ and $Y'$ are allowed to have the same topology and $Y=Y'=\emptyset$. – Mirko Jul 24 '19 at 04:41