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Suppose $X$ is a compact Hausdorff space, $Y$ is a compact $T_1$ space and $f\colon X\to Y$ has the property that for every $T_3$ space $Z$ and every continuous $g\colon Y\to Z$, the composition $g\circ f$ is continuous.

Can we conclude that $f$ itself is continuous?

If $Y$ was allowed to be more pathological, the answer would no: for example, if we take for $Y$ the Sierpiński two-point space (which is compact and $T_0$), then any continuous function from $Y$ to a $T_3$ (even $T_1$) space is constant, so the assumption is trivially satisfied for any function $f$, which means that in particular, we can take an $f$ which maps a non-closed set to the closed point, and its complement to the open point, and then the conclusion fails.

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    What is your definition of $T_3$? (as far as I know topology books differ on that matter) .... since I am usually working with the one, s.t. $T_3 \Rightarrow T_1$, i.e. we could take $Z:= Y$ and $g := id_Y$? – 2017-01-10
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    Depends on what you mean by $T_3$ space. Wikipedia says that a $T_3$ space is "usually" defined as a Hausdorff space that satisfies the $T_3$ axiom. If so, then you can take $Y$ to be the co-finite topology. Then you again get that, any $Y\to Z$ with $Z$ $T_3$ must be constant. If a $T_3$ space is not necessarily Hausdorff, that might give you more functions. – 2017-01-10
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    $Z=Y$ is not allowed, because $Y$ is fixed, and $Y$ is $T_1$, not necessarily $T_3$. @M.U. – 2017-01-10
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    Of course, you are right. An embarrassing "mind twist". – 2017-01-10
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    By $T_3$ i mean regular Haudorff, i.e. you can separate a closed set from a point with open sets. In this case, you can just take $Z$ to be compact Hausdorff, but I am also curious in more general case when $X$ and $Y$ need not be compact. – 2017-01-10

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The answer is no.

Consider $X=[0,1]$ and $Y=([-1,1]\times \{0,1\})/{\sim}$, where $(x,i)\sim (x',i')$ iff $(x,i)=(x',i')$ or $x=x'<0$ (i.e. $Y$ is a double-ended interval with "loose threads" hanging off the twin ends). Put $f\colon X\to Y$ as $0\mapsto (0,1)$, and $x\mapsto (x,0)$ when $x>0$.

Then $f$ is clearly not continuous at $0$, but if we compose it with any function into a Hausdorff space, we will obtain a continuous function (because any continuous function from $Y$ to a Hausdorff space will glue $(0,0)$ and $(0,1)$ together).