Whether Hausdorff Property is Required 
Let $H$ denote a group that is also a Hausdorff topological space. Show that $H$ is a topological group if and only if the map $h: H \times H \rightarrow H, h:=(x,y)\rightarrow xy^{−1}$ is continuous.

The following proof does not use the Hausdorff property.
Suppose $h$ is continuous. 
$f:H \rightarrow H \times H, f(x):=(e, x)$ is continuous (Munkres theorem 19.6). $(h\circ f)(x)=x^{−1}$ is continuous. 
$j:H\times H \rightarrow H \times H, j(x, y):=(x, y^{−1})$. $\pi _1 (x, y)=x$ is continuous. $(h\circ f \circ \pi _2)(x, y)=y^{−1}$ is continuous. By Munkres theorem 19.6, $j$ is continuous. $(h\circ j)(x, y)=xy$ is continuous.
So $H$ is a topological group.
Suppose $H$ is a topological group. $k(x, y):=xy$ and $l(x):=x^{−1}$. 
$\pi _1 (x, y)=x$ is continuous. $(l \circ \pi _2) (x, y)=x^{−1}$ is continuous. By Munkres theorem 19.6, $j$ is continuous. $j \circ k=h$ is continuous.
Please let me know whether Hausdorff property is required. (I'm aware that a topological group is Hausdorff).
Thanks.
Edit: Thanks to Aryaman Maithani's comment, some textbooks (for eg Munkres) require Hausdorff-ness in the  definition of topological group. Hausdorff-ness is required probably only in that case.

Indeed the proof does not require any Hausdorffness, we only need that certain other natural product maps are continuous, and this holds regardless of separation axioms.
I would do it this way: suppose $h$ is continuous, then indeed $f:x \to (e,x)$ from $H$ into $H \times H$ is continuous (the composition with the two projections are a constant map (always continuous) resp the identity on $H$ (ditto). The inversion map $I:x \to x^{-1}: H \to H$ then is continuous as $h \circ f$, a composition of continuous maps. The product map $P: (x,y) \to xy$ is then also continuous as $h \circ (1_H \times I)$ etc.
And if $P$ and $I$ are continuous so is $h$ as the composition  $P \circ (1_H \times I)$, so both implications hold.
